continuation of fix A to fix B ?

[JmsNxn]

correct.

  I never thought I'd see the day where Tommy just agrees with me! Without some extra comments or something  

Thanks, Tommy... I'd be willing to post more about Hilbert spaces, but I'm pretty nervous that it'd go on deaf ears in this forum.

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For example, if we take:

\[
F_{\lambda} (s) = \lim_{n\to\infty} \log_{\sqrt{2}}^{\circ n} \beta_{\lambda}(s+n)\\
\]

Where: \(\beta_{\lambda}(s+1) =\sqrt{2}^{\beta_{\lambda}(s)} + O(e^{-\lambda s})\).

Then, if \(\lambda \in \mathbb{R}^+\) and \(\lambda > -\log\log(2)\), we should have the ability to write, for \(0 < \Im(s) < \frac{2\pi}{\lambda}\) the following transform:

\[
\mathcal{H}_\lambda(\xi) = \int_{-\infty}^\infty F_{\lambda}(s) e^{-2\pi i \xi s}\,ds\\
\]

We will take this transform as a line integral in the complex plane, but only in the strip \(0 < \Im(s) < \frac{2\pi}{\lambda}\).

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What I Kind of believe Tommy has figured out, is that \(\lambda\) and \(\xi\) will share a relationship. So let's say fuck it, and just set \(\lambda = -\log\log 2\). This means that \(F(s)\) is just the standard super function, we can even renormalize it if you want, which just equates to multiply \(\mathcal{H}(\xi)\) by \(e^{2 \pi i s_0 \xi}\). But for the sake of the post, we will write:

\[
\mathcal{H}(\xi) = \int_{-\infty}^\infty F(s) e^{-2\pi i \xi s}\,ds\\
\]

Where \(F(0) = 1\) and \(F: (-2,\infty) \to \mathbb{R}\) bijectively, and \(F(s+1) = \sqrt{2}^{F(s)}\), and this is the regular iteration (Schroder iteration).

This \(F\) satisfies being holomorphic for \(0 < \Im(s) < -\frac{2\pi}{\log\log 2}\), and \(F(-\infty) = 4\) and \(F(\infty) = 2\) on this strip. So a clearer way of writing this integral, just to be technical, is take \(0 < t < -\frac{2\pi}{\log\log 2}\):

\[
\mathcal{H}(\xi) = \int_{-\infty+it}^{\infty + it} F(s) e^{-2\pi i \xi s}\,ds\\
\]

The value of this integral is unchanged by choice of \(t\) (so long as it's in the strip).

There is a specific domain in \(\xi \in \mathbb{C}\) where this Fourier transform is convergent. Finding it is a little tricky, but not really. We are literally just writing \(c = e^{-2\pi i \xi}\) from the discussion above, and everything is the same. The trouble is, we have to normalize this in Tommy's language, so that \(F(\infty) = 0\), this can be accomplished by just writing \(\widetilde{F}(s) = F(s) - 2\), and then everything's much better (but technically the same in distribution analysis). Then we write:

\[
\widehat{F}(\xi) =  \int_{-\infty+it}^{\infty + it} (F(s)-2) e^{-2\pi i \xi s}\,ds\\
\]

The value \(\widehat{F}(\xi)\) is for all intents and purposes a Fourier Transform on the iterates... And this object will converge, and it will converge absolutely. For all \(e^{-2\pi i \xi} = c\) where \(1 < |c| < \log(2)\). Same as I wrote above...

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We can effectively encode in this manner--the action of:

\[
F(s+h) = \exp_{\sqrt{2}}^{\circ h} F(s)\\
\]

Into:

\[
\widehat{\exp_{\sqrt{2}}^{\circ h} F}(\xi)= e^{2 \pi i h \xi}\widehat{F}(\xi)\\
\]

And we've "fourier linearized" as opposed to typical linearizations. This is a much more difficult thing to accomplish. This is groundbreaking, and I think Tommy has shone the light on the final key!

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THIS ONLY HAPPENS IN VERY SPECIAL CASES! This can get really tricky with general iteration. But for something like \(\sqrt{2}\), we don't have to worry  !

This is a lot of what I'm struggling with right now. But I believe we can write every solution to tetration base \(\sqrt{2}\) in this manner, by multiplying \(\widehat{F}\) by a periodic solution, and doing Hilbert space magic. I'm not there yet, and I don't really post a lot of this here. Largely because this isn't the forum for that!

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Deepest Regards, James

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I should add that my switch from \(F\) to \(F-2\) is entirely rigorous, because in distributional analysis/advanced fourier analysis, constants can be Fourier transformed. So for example, in advanced scenarios:

\[
\int_{-\infty}^\infty e^{-2\pi i \xi s}\,ds
\]

Is a valid expression on its own, and that \(F\) and \(F-2\) are only seperated by a distributional function. So to treat them similarly is more of an abuse of notation than an error. I used \(F-2\) and called the fourier transform an action on this, just to shorten the discussion

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Final edit, but:

\[
f^{\circ s}(z) = F(s+ a(z))\\
\]

And therefore:

\[
\widehat{F(s+a(z))} = e^{2\pi i \xi a(z)} \widehat{F}(\xi)\\
\]

Upon which, as \(F(s) = f^{\circ s}(1)\), we can equally define \(f^{\circ s}(z)\) for \(z\) in the appropriate domains. And this just equates to multiplication by a fourier style exponential....

LET'S FUCKING GO!