A first hires look at tetration \(\lambda = 1\) and \(b = e\)

[JmsNxn]

I've been making some different graphs of \(\beta\) and I got a good one to share.

...
All of these will be committed towards the asymptotic thesis of the beta function. Which is that the beta method approaches at least an asymptotic expansion at each point, as opposed to a Taylor series. This is compatible with everything I have been saying, and additionally compatible with Sheldon's work. This paper will entirely focus on ASYMPTOTIC behaviour. Which looks like tetration; but if you try and make it tetration, expect a good amount of errors.

very nice.  I spent a bit more time on b=sqrt(2), and I believe it converges it can be proven to converge analytically so long as the imaginary period of lambda is less than the imaginary period of the attracting fixed point=2.  So lambda=1 would be analytic, but lambda=0.3 would not converge, since
\(\frac{2\pi i}{0.3}>\frac{2\pi i}{\ln(\ln(2))}\)

Success!!!

I've been able to prove this unequivocally. I'll write a quick sketch.

To begin, define:

\[
\beta_{\lambda,\mu}(s) = \Omega_{j=1}^\infty \frac{e^{\mu z}}{1+e^{\lambda (j-s)}}\,\bullet z\\
\]

And for simplicity, we'll stick to \(\mu = \log(2)/2\) and let \(\lambda\) vary. We want to show that for \(\Re \lambda \le -\log \log 2\) the iterated \(\log\)'s will no longer converge. To start, take our sequence of approximations:

\[
\tau^{n+1}(s) = \log(1+\frac{\tau^n(s+1)}{\beta(s+1)}) - \log(1+e^{-\lambda s})\\
\]

Where the \(\log\)'s are base \(e^{\mu}\). I can show the following result if requested; but each:

\[
\tau^{n}(s+k) = \mathcal{O}(e^{-\lambda k})\\
\]

And consequently, by looking at the above equation; and rearranging; for \(n \ge 2\) we must have:

\[
\frac{\tau^n(s+k)}{\mu\beta(s+k)} = \mathcal{O}(e^{-\lambda k})\\
\]

And again, by the above asymptotic,

\[
\frac{1}{\mu\beta(s+k)} = \mathcal{O}(1)\\
\]

And now, let's call this bound \(A_\mu\). For the case of \(\mu = \log(2)/2\), we get the bound \(A_\mu = 1/\log(2)\). Now, since all of these things decay exponentially; we can linearize all of this discussion. So everywhere you have a \(\log(1+x)\), replace it with \(\frac{x}{\mu}\) which are asymptotic equivalents as \(x \to 0\). From here we can make a new \(\tau\), let's call it \(\widetilde{\tau}\), which effectively just removes the \(\log\)'s with this asymptotic.

To reconcile:

\[
\limsup_{k\to\infty}\left|\frac{1}{\mu \beta(s+k+1)}\right| = A_\mu\\
\]

\[
\widetilde{\tau}^{n+1}(s) = \frac{\widetilde{\tau}^n(s+1)}{\mu \beta(s+1)} - \frac{e^{-\lambda s}}{\mu}\\
\]

Which has the closed form expression:

\[
\widetilde{\tau}^{n+1}(s) = -\sum_{j=0}^n \frac{e^{-\lambda(s+j)}}{\mu^{j+1} \prod_{c=1}^j \beta(s+c)}\\
\]

Now, necessarily:

\[
\left|\frac{e^{-\lambda(s+j)}}{\mu^{j+1} \prod_{c=1}^j \beta(s+c)}\right| \le e^{-\Re(\lambda s) + (\log A_\mu - \Re\lambda)j}\\
\]

Where we've let \(\Re(s)\) be arbitrarily large for the bound to come in effect.  And voila, this series only converges if \(\Re\lambda > \Re \log A_\mu\), and for \(\mu = \log(2) / 2\), the value \(\Re \log A_\mu = -\log \log (2)\)!!!!!!!!!!!!


You will see something very similar everywhere in the Shell-Thron region; where essentially \(A_\mu = \frac{1}{\mu \omega}\), where \(\omega\) is the fixed point. This will cause the \(\tau\) sequence to be very irregular as soon as \(\Re \lambda \le -\Re \log \mu \omega\). But at \(\lambda = -\log \mu \omega\) is precisely when we achieve the period of the standard Schroeder iteration!

Makes so much sense now!

This is just a quick sketch at the moment, but for the \(\sqrt{2}\) case \(\beta\) has really nice behaviour; so it's a proof for this case; for other cases you have to be more careful when talking about domains; especially when you make this change to a linear approximation.

Regards, James

Now, to solve this for \(\Re \lambda < \log \log (2)\), we have to introduce convergents \(p_n\) to make this sum converge; still not certain how to do this effectively as of yet. I won't go into too much detail; but I now understand why base \(e\) and \(\lambda = 1\) results in no where analytic on the real line. This is harder to prove; but is god damned fascinating to think about.