Rank-Wise Approximations of Hyper-Operations

[JmsNxn]

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 \(a=e\).

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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...

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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.