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Rank-Wise Approximations of Hyper-Operations
#1
Question 
When [Image: png.image?\dpi%7B110%7D%20a\uparrow\uparrow%20x] converges and a does not equal one, you could possibly extend tetration at base a by using the limit formula here.
Then you could possibly use the same limit formula to extend pentation, hexation et cetera at base a.
Does anyone know of any approximations of these hyper-operations that become more and more accurate as the ranks of the hyper-operations increase?
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#2
(06/04/2022, 01:55 AM)Catullus Wrote: When a^^x converges and a does not equal one, you could possibly extend tetration by using the limit formula here: https://math.eretrandre.org/tetrationfor...42#pid4442. Then you could use the same formula to extend pentation, hexation, et cetera for those bases. Does anyone know of any approximations of those hyper operators that become more and more accurate as the rank of the hyper operation increases?

Hey!

So this formula produces the standard Schroder iteration, or the standard Abel iteration.

I don't think this would work for \(a>\eta=e^{1/e}\), if it does, it would probably look like Kouznetsov's approach to Kneser. But I can 90% guarantee it wont work on the real line.

I can prove this is the standard Schroder/Abel iteration if you like.

For \(a > \eta\) you would have to focus about a fixed point \(L \in \mathbb{C}\), and then you are just performing Schroder about \(L\). This would not produce a real valued tetration about \(a=e\).


So, technically you are correct. This formula would produce tetration! And it would produce it pretty much everywhere! But, it's the bottom of the barrel for repelling cases, it produces non real values on the real positives. We can't have e^^1/2 being a non-real number...



That is a cool as fuck formula though, lol. Never noticed it before.
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#3
(06/04/2022, 02:05 AM)JmsNxn Wrote:
(06/04/2022, 01:55 AM)Catullus Wrote: When a^^x converges and a does not equal one, you could possibly extend tetration by using the limit formula here: https://math.eretrandre.org/tetrationfor...42#pid4442. Then you could use the same formula to extend pentation, hexation, et cetera for those bases. Does anyone know of any approximations of those hyper operators that become more and more accurate as the rank of the hyper operation increases?

Hey!

So this formula produces the standard Schroder iteration, or the standard Abel iteration.

I don't think this would work for \(a>\eta=e^{1/e}\), if it does, it would probably look like Kouznetsov's approach to Kneser. But I can 90% guarantee it wont work on the real line.

I can prove this is the standard Schroder/Abel iteration if you like.

For \(a > \eta\) you would have to focus about a fixed point \(L \in \mathbb{C}\), and then you are just performing Schroder about \(L\). This would not produce a real valued tetration about \(a=e\).


So, technically you are correct. This formula would produce tetration! And it would produce it pretty much everywhere! But, it's the bottom of the barrel for repelling cases, it produces non real values on the real positives. We can't have e^^1/2 being a non-real number...



That is a cool as **** formula though, lol. Never noticed it before.

I said tetrations of a would converge. a would not be e.
Also why must e^^1/2 be a real number? The Bennet interpolation of the Fibonacci numbers gives a non real number for the halfth Fibonacci number. The tetration being non real valued seems much cooler!
Can you please tell me the proof that this is the standard Schröder/Abel iteration. Also the Abel iteration gives a real number for e^^1/2.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#4
(06/04/2022, 02:14 AM)Catullus Wrote:
(06/04/2022, 02:05 AM)JmsNxn Wrote:
(06/04/2022, 01:55 AM)Catullus Wrote: When a^^x converges and a does not equal one, you could possibly extend tetration by using the limit formula here: https://math.eretrandre.org/tetrationfor...42#pid4442. Then you could use the same formula to extend pentation, hexation, et cetera for those bases. Does anyone know of any approximations of those hyper operators that become more and more accurate as the rank of the hyper operation increases?

Hey!

So this formula produces the standard Schroder iteration, or the standard Abel iteration.

I don't think this would work for \(a>\eta=e^{1/e}\), if it does, it would probably look like Kouznetsov's approach to Kneser. But I can 90% guarantee it wont work on the real line.

I can prove this is the standard Schroder/Abel iteration if you like.

For \(a > \eta\) you would have to focus about a fixed point \(L \in \mathbb{C}\), and then you are just performing Schroder about \(L\). This would not produce a real valued tetration about \(a=e\).


So, technically you are correct. This formula would produce tetration! And it would produce it pretty much everywhere! But, it's the bottom of the barrel for repelling cases, it produces non real values on the real positives. We can't have e^^1/2 being a non-real number...



That is a cool as **** formula though, lol. Never noticed it before.

I said tetrations of a would converge. a would not be e.
Also why must e^^1/2 be a real number? The Bennet interpolation of the Fibonacci numbers gives a non real number for the halfth Fibonacci number. The tetration being non real valued seems much cooler!
Can you please tell me the proof that this is the standard Schröder/Abel iteration. Also the Abel iteration gives a real number for e^^1/2.

Hey!

Okay, this will be a bit long winded...

First of all, a^^z is always solvable. It's which solution that matters. Why would you choose a^^1/2 is complex when you can find one that's real. This doesn't mean one is better than the other; this means we are searching for different criteria. This forum is not about the one and only solution of tetration; it's about the many ways to solve tetration. So yes, what is written here is a solution. It's a solution solved by Schroder though, and ultimately is not that helpful, other than with iterated exponentials near a complex fixed point \(a^L = L\). This isn't a problem. I was choosing \(e\) as an example of where all of this would fail. Any \(a>\eta\) suffers the problems this has when \(a=e\).


Now, in the care of the post you referenced. The iterated function is \(f^{\circ s}(\xi)\), where \(\xi\) is near an attracting or repelling fixed point \(\xi_0\) (or neutral attracting). This is stated in the disclaimer of the post. Without loss of generality we can assume it is attracting or neutral attracting. Now if I take the Schroder iteration, or the Abel iteration (if it's neutral attracting), F. Then:

$$
F(s,\xi)\,\,\text{is holomorphic for}\,\xi\,\text{in a petal about}\,\xi_0\\
$$

The values:

$$
F(k)\\
$$

Satisfy the functional equation from the post you referenced. Therefore the limit of the functional equation you posted satisfies the equation just like \(F(k)\).

Enter Ramanujan's identity theorem.

Any function \(F(s)\) which is less than \(e^{\rho\Re(s) + \tau|\Im(s)|}\) for \(\tau < \pi/2\) has an identity theorem on the naturals \(\mathbb{N}\). So the Schroder iteration \(F(s)\) satisfies these bounds. And it satisfies the linked identity. But additionally, a net of values about the limit suggested in this formula converges to \(F\), because each value at \(F(k)\) is unique.

I apologize if this is quick and to the point. But this is absolutely Schroder's iteration. Essentially it's just a different way of writing it.
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#5
(06/04/2022, 04:20 AM)JmsNxn Wrote: First of all, a^^z is always solvable. It's which solution that matters. Why would you choose a^^1/2 is complex when you can find one that's real. This doesn't mean one is better than the other; this means we are searching for different criteria. This forum is not about the one and only solution of tetration; it's about the many ways to solve tetration. So yes, what is written here is a solution. It's a solution solved by Schroder though, and ultimately is not that helpful, other than with iterated exponentials near a complex fixed point \(a^L = L\). This isn't a problem. I was choosing \(e\) as an example of where all of this would fail. Any \(a>\eta\) suffers the problems this has when \(a=e\).
A real valued tetration like that would not agree with the limit formula from https://math.eretrandre.org/tetrationfor...42#pid4442. However there might be a complex valued one that does! How would that tetration behave? Also the main point of this thread was going to be about approximation(s) of convergent hyperoperations that become better and better as the rank goes up, like how when 1<a<η a^^x can be approximated better and better as x becomes bigger and bigger by LambertW(-ln(a))/ln(a)-b*LambertW(-ln(a))^x. Where b=lim x → ∞ (LambertW(-ln(a))/ln(a)-a^^x)/LambertW(-ln(a))^x. It might be that as x grows larger and larger for some a a[x]b converges exponentially to a for any b. sqrt(2)+2~3.414. sqrt(2)*2~2.828. sqrt(2)^2=2. sqrt(2)^^2~1.633. sqrt(2)^^^2~1.520. They do seem to converge. How would we approximate the convergence as x grows larger and larger of a[x]b as a function continuous in terms of the x variable that becomes better and better as x grows larger and larger, when a[x]b converges? Even if a>η (Resulting in a non-real valued tetration.) a[x]b, for any b may converge with larger and larger x.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#6
(06/04/2022, 04:27 AM)Catullus Wrote:
(06/04/2022, 04:20 AM)JmsNxn Wrote: First of all, a^^z is always solvable. It's which solution that matters. Why would you choose a^^1/2 is complex when you can find one that's real. This doesn't mean one is better than the other; this means we are searching for different criteria. This forum is not about the one and only solution of tetration; it's about the many ways to solve tetration. So yes, what is written here is a solution. It's a solution solved by Schroder though, and ultimately is not that helpful, other than with iterated exponentials near a complex fixed point \(a^L = L\). This isn't a problem. I was choosing \(e\) as an example of where all of this would fail. Any \(a>\eta\) suffers the problems this has when \(a=e\).
A real valued tetration like that would not agree with the limit formula from https://math.eretrandre.org/tetrationfor...42#pid4442. However there might be a complex valued one that does! How would that tetration behave? Also the main point of this thread was going to be about approximation(s) of convergent hyperoperations that become better and better as the rank goes up, like how when 1<a<η a^^x can be approximated better and better as x becomes bigger and bigger by LambertW(-ln(a))/ln(a)-b*LambertW(-ln(a))^x. Where b=lim k→∞ (LambertW(-ln(a))/ln(a)-a^^x)/LambertW(-ln(a))^x. It might be that as x grows larger and larger for some a a[x]b converges exponentially to a for any b. sqrt(2)+2~3.414. sqrt(2)*2~2.828. sqrt(2)^2=2. sqrt(2)^^2~1.633. sqrt(2)^^^2~1.520. They do seem to converge. How would we approximate the convergence as x grows larger and larger of a[x]b as a function continuous in terms of the x variable that becomes better and better as x grows larger and larger, when a[x]b converges? Even if a>η (Resulting in a non-real valued tetration.) a[x]b, for any b may converge with larger and larger x.

Hey,

So it's known that:

$$
\lim_{n\to\infty} \sqrt{2} \uparrow^n x \to \sqrt{2}\\
$$

As \(n\to\infty\) and \(x > 1\) (it gets a little wonky for \(0 < x <1\)). This is actually the basis of my fractional calculus approach to semi-operators. This is assuming we are taking the standard Schroder iteration.

If you take \(1 \le \alpha \le \eta\):

$$
\alpha \uparrow^n z \to \alpha\\
$$

For \(\Re(z) >1-\delta\).

Now, the idea would be to approximate this limit (it converges geometrically, was never able to make an estimate of how fast, geometrically). Then we'd want to interpolate:

$$
\vartheta_n(w,z) = \sum_{k=0}^\infty \left(\alpha \uparrow^{n+k} z\right) \frac{w^k}{k!} \to \alpha e^{w}\,\,\text{as}\,\,n\to\infty\\
$$

Then, the fractional iteration:

$$
\frac{d^s}{dw^s}\Big{|}_{w=0} \vartheta_n(w,z) = \alpha \uparrow^{n+s} z\\
$$

It's very difficult to numerically evaluate, but many arguments were showing it should converge. I hit a brick wall though which stopped me from continuing this approach.


And as I said, the complex tetration that agrees with this limit is the standard Schroder iteration about which ever fixed point you want.

So if we take this limit about \(L\) where \(e^L = L\), then the Schroder function satisfies:

$$
\begin{align*}
\Psi(e^z) &= e^L \cdot \Psi(z)\\
\Psi(L) &= 0\\
\end{align*}
$$

Then, inverting this function to get \(\Psi^{-1}\), which is entire, there exists countably infinite \(s_0 \in \mathbb{C}\), such that:

$$
\text{tet}(s) = \Psi^{-1}(e^{L(s-s_0)})\\
$$

This iteration will equal the one in the link you suggested, provided what \(s_0\) we choose.

The trouble then becomes, which value of \(s_0\) and which fixed point, but there are countable solutions to tetration. None will be real valued though.


If you view this as an iteration instead, this solves for:

$$
\exp^{\circ s}(z)\\
$$

When \(|z-L| < \delta\), for what ever \(L\) you want. You can actually identify this from the limit formula. The parameter \(n\) is only used in the exponent of \(f'(u)\) where \(u\) is close to \(L\). We can effectively set this to \(L\), as it's only getting closer (wlog because repelling/attracting about a fixed point is the same thing--can be treated as attracting). So the exponent looks like \(e^{Ln}\). This has a period \(2\pi i/L\) in \(n\), and there is only one iteration per period, and since the Schroder iteration has this period, it's just the Schroder iteration.


It's a really nifty formula, but I'm not sure how it enlightening it is to the general theory.
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#7
(06/05/2022, 11:34 PM)JmsNxn Wrote: The trouble then becomes, which value of \(s_0\) and which fixed point, but there are countable solutions to tetration. None will be real valued though.
They should all produce the same Shröder iteration.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#8
(06/06/2022, 03:36 AM)Catullus Wrote:
(06/05/2022, 11:34 PM)JmsNxn Wrote: The trouble then becomes, which value of \(s_0\) and which fixed point, but there are countable solutions to tetration. None will be real valued though.
They should all produce the same Shröder iteration.

Hmm, I'm not sure what you mean here. They all spawn from the same Schroder iteration (I'm using this word a little differently than I guess how it should be used).

So,

$$
\exp^{\circ s}(z) = \Psi^{-1}(e^{Ls}\Psi(z))\\
$$

Is the one and only Schroder iteration about \(z \approx L\). What I was saying, is that, there exists countable \(s_0\) such that:

$$
\Psi^{-1}(e^{-Ls_0}) = 1\\
$$

So, there are countable functions:

$$
F(s) = \Psi^{-1}(e^{L(s-s_0)})\\
$$

Such that:

$$
\begin{align}
F(0) = 1\\
F(s+1) = e^{F(s)}\\
F(s+2\pi i/L) = F(s)\\
\end{align}
$$

But there is only one Schroder iteration of the exponential. There are countable TETRATIONS which satisfy this. I apologize if I talked too loosely.

Just a quick note, on this forum a Schroder iteration will always refer to standard/regular iteration about a fixed point. BUT Tetration is to me something that satisfies the functional equation and an initial condition of \(F(0)=1\). So the standard Schroder iteration of \(e^z\) about \(L\) creates countably many tetrations....
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#9
(06/06/2022, 04:00 AM)JmsNxn Wrote: But there is only one Schroder iteration of the exponential. There are countable TETRATIONS which satisfy this. I apologize if I talked too loosely.
Tetraion is iterated exponentiation.
Using the limit formula on any of the fixed points should produce the same tetration function.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#10
(06/06/2022, 04:07 AM)Catullus Wrote:
(06/06/2022, 04:00 AM)JmsNxn Wrote: But there is only one Schroder iteration of the exponential. There are countable TETRATIONS which satisfy this. I apologize if I talked too loosely.
Tetraion is iterated exponentiation.
Using the limit formula on any of the fixed points should produce the same tetration function.

No, that's incorrect.

I just explained why that's incorrect.

A tetration function:

$$
\text{Tet}(s)\\
$$

Is a function such that:

$$
\begin{align}
\text{Tet}(0) = 1\\
\text{Tet}(s+1) = e^{\text{Tet(s)}}\\
\end{align}
$$

As I just showed, there are countable solutions to this equation when you use Schroder's iteration.

To reiterate. The inverse Schroder function about \(L\) of \(e^z\) is an entire function, \(\Psi^{-1}(z)\). This function equals \(1\) countably infinite times (Picard). Therefore, \(\Psi^{-1}(e^{-Ls_0})\) equals \(1\) countably infinite times. Therefore \(\Psi^{-1}(e^{L(s-s_0)})\) is a tetration function. Therefore there are countably infinite tetration functions spawned from the fixed point \(L\) using Schroders iteration.

All I am doing is filtering through preimages.
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