Rank-Wise Approximations of Hyper-Operations

[JmsNxn]

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.