continuation of fix A to fix B ?

[tommy1729]

Yes, Tommy!

That's very much my point. This finds the complimentary solution. I think we can work a tad differently though; so let me take your example but flip it slightly on its head. Let:

\[
f(z) = 5z + z^3\\
\]

Now define the function:

\[
\gamma_c(w) = \Omega_{j=1}^\infty \frac{w f(z)}{w+c^j}\,\bullet z\\
\]

For \((|c| > 1\). This function is holomorphic in \(c\) and \(w\), so long as \(w \neq -c^j\) for \(j \ge 1\). And further, is a meromorphic function \(\gamma_c(w) : \mathbb{C}^2 \to \widehat{C}\). This function satisfies:

\[
\gamma_c(cw) = \frac{w \gamma_c(w)}{w+1}\\
\]

Now, we can take the iteration:

\[
\Psi_c(w) = \lim_{k\to\infty} f^{-k} \gamma(c^k w)\\
\]

This iteration only converges for \(|c| > 5\).

My point is that, if we look at, instead, your summation condition, (like the one you wrote), we should be able to solve if for \(1 < |c| < 5\)...

The uniqueness of these equations is difficult to describe--and it requires understanding the uniqueness of these almost elliptic functions \(\theta(z+1) = \theta(z)\) and \(\theta(z+2 \pi i/\log( c)) = \theta(z) - 2\pi i/\log(c )\). These functions are unique UPTO an elliptic function and the location of the singularities and their singular parts.

Suppose:

\[
\theta_1 \neq \theta_2\\
\]

Suppose every where \(\theta_1\) has a singularity \(z_0\), so does \(\theta_2\). Suppose additionally, the singular parts of this singularity are equivalent (this is difficult to describe properly, so I'll use a weaker condition at the moment):

\[
\lim_{z \to z_0}|\theta_1(z) - \theta_2(z)| < \infty\\
\]

Then, \(\theta_1 = \theta_2 + C\) for a constant \(C\), because:

\[
\theta_1(z) - \theta_2(z) = \wp(z)
\]

And:

\[
\wp(z+1) = \wp(z)\,\,\,\,\wp(z+2\pi i/\log( c)) = \wp(z)\\
\]

The function \(\wp(z)\) has no singularities. Therefore it is bounded on \(\mathbb{C}\). By Liouville's theorem, therefore it is constant.

So essentially, any solution to these types of equations, only differ by the singular behaviour of their theta function... Which equates to the statement that they are equivalent upto an elliptic function!

Yes absolutely.

Double periodic functions usually have addition formula.

I mentioned your remark in another recent thread : https://math.eretrandre.org/tetrationfor...p?tid=1690

The hunt for cases where our double periodic is just id(z) is open !

AS some test cases I propose computing supers abels and half-iterates by the methods above for 

f(x) = x^3

f(x) = x^9

( will the half iterate be x^3 ???!!!??? )

and the case where the derivatives are interesting as discovered by Bo :

ITERATION WITH 2 ANALYTIC FIXED POINTS

Iteration wit

 https://math.eretrandre.org/tetrationfor...605&page=4

post 33 :

Ok, I even can give a polynomial
\[ p (x)  = x ^ 3 + \frac{\sqrt{5}-3}{2} x^2 - \frac{\sqrt{5}-3}{2} x \]

has fixed points 0 and 1 and 
\[ p'(x) = 3x^2 + (\sqrt{5}-3)x - \frac{\sqrt{5}-3}{2} \]

The derivatives at the fixed points are:
\( p'(0) = -\frac{\sqrt{5}-3}{2}, p'(1) = 3 + \frac{\sqrt{5}-3}{2}=\frac{\sqrt{5}+3}{2}\) hence 
\[ p'(0)p'(1) = -\frac{(\sqrt{5}-3)(\sqrt{5}+3)}{4} = 1\]

****

when picking an approp c this might get interesting.

The fact that p has a similar period at both fixpoints makes the suggestion that this might have a large domain of analyticity ?!

The polynomials x^3 and x^9 have a flat derivative at 0 but the method still works.

Will we get id(z) ? x^sqrt(3) ?

and if not , how close will we get ?

Really interesting stuff.

A possible connection with addition formula is in my mind.

As always I have many many more ideas and conjectures but this will do for today.

regards

tommy1729

[JmsNxn]

Let's let \(f(x) = x^9\), then the fractional iteration is precisely \(f^{\circ s}(x) = x^{9^s}\)--whereby \(f^{\circ \frac{1}{2}}(x) = x^3\).

I have not investigated these cases at all. These relate to the cases \(f'(0) = 0\)--and we have to use the Bottcher coordinate. I have never investigated infinite compositions as they relate to Bottcher's coordinates--as they become much more volatile. \(f^{\circ s}(x)\) is holomorphic not holomorphic in a neighborhood of \(x=0\) (generally), unless we invoke a natural iterate.

So, for example, if we write:

\[
\gamma_c(w) = \Omega_{j=1}^\infty \frac{wf(z)}{w+c^j}\bullet z\\
\]

Then:

\[
\gamma_c(cw) = \frac{w f(\gamma_c(w))}{w+1}\\
\]

I highly doubt, choosing \(f^{-1}(z) = \sqrt[9]{z}\), that:

\[
\lim_{k\to\infty} f^{-k} \gamma_c(c^kw)\\
\]

Converges for any \(c\). This is based on the heuristic, when \(|f'(0)| = 1\), this object converges for \(|c| > 1\), and for \(|f'(0)| = a > 1\), that this object only converges for \(|c| > a\). When \(|f'(0)| = a < 1\), then this limit converges for: \(|1/c| < a\). So when \(a =0\), we can expect \(|1/c| < 0\)--which means it should never converge. Though of course, there may be some boundary behaviour. We may be able to get an abel function, but we'd have to finesse a lot.

I haven't done any research on the infinite composition method when dealing with \(f'(0) = 0\) and the Bottcher coordinate. So I can't say forsure. It's never interested me, as this opens us up to a lot of branching talk right off the get go...... And it appears pretty naturally that Mock-Schroder is useless.

We could probably make a mock Bottcher though...

So we maybe able to solve:

\[
H(w^9) = f(H(w))h(w)\\
\]

Which would be written as:

\[
H(w) = \Omega_{j=1}^\infty h(w^{9^{-j}})f(z)\bullet z\\
\]

This may be able to approximate the standard Bottcher coordinate, considering \(h\) is well enough behaved... We would write the limit as:

\[
f^{-j} H(w^{9^j}) \to \Psi\\
\]

Where:

\[
\Psi(w^9) = f(\Psi(w))\\
\]

Which may converge to a Bottcher coordinate, or the standard one.--which would allow us to reconstruct the standard fractional iterate \(f^{\circ s}(x) = x^{9^s}\)...

But since \(f(x) = x^9\) is its own Bottcher coordinate, we'd have that \(\Psi(w) \to w\).... this seems unlikely, or at least difficult to pull off. It could happen though, I don't see why not. I hate Bottcher coordinates, so I've done next to no research on the super attracting case of dynamics. Infinite compositions can definitely aid, but I don't think they shed as much light as in the Schroder/Abel case--geometric/parabolic...

For example, setting:

\[
h(w) = w-1\\
\]

Then \(H(w)\) is holomorphic everywhere:

\[
\sum_{j=1}^\infty |w^{9^{-j}} -1| < \infty\\
\]

Which is for \(w \in \mathbb{C}/(-\infty,0)\). So that:

\[
H(w^9) = H(w)^9 (w-1)\\
\]

Maybe the iterated thing would converge to \(\Psi(w) = w\)... would be an interesting test to run! If it does, it should follow that it works for more complex beasts, where \(f(x) = x^9 + O(x^{10})\). Or for any general \(f(x) = x^n + O(x^{n+1})\).... And reducing to the Bottcher's coordinate in the general sense. Unfortunately, Bottcher's coordinate, doesn't allow for as many natural generalizations. Schroder's generalizations, allow us to create arbitrary multipliers in the Schroder function. Any function \(f(w^9) = f(w)^n\), is well understood and more restrictive. So I can't imagine us having as much freedom.

---

EDIT:

But, using your formula, we look at:

\[
P_c(x) = \sum_{k=-\infty}^\infty c^k x^{9^k}\\
\]

For \(|c| > 1\) and \(|x| < 1\) and \(x\not \in (-1,0)\), then, absolutely this thing is holomorphic, by which:

\[
P_c(x^9) = cP_c(x)\\
\]

This kind of confirms my suspicion, that this is the complement of the infinite composition/mock schroder/levenstein approach. Where everything fails in my case, your method picks up the slack entirely. And where my method usually works, yours fails, and where mine fails, yours works. And this seems to be universally happening!

Very interesting regardless!!!!

So for instance, using Neutral fixed points \(f'(0) = 1\), I expect your formula NEVER CONVERGES for \(|c|>1\), because my formula converges EVERYWHERE \(|c|>1\). For Geometric fixed points, \(|f'(0)| \neq 0,1\), my formula converges where yours doesnt, and yours converges where mine doesn't. And for super attractive \(f'(0) = 0\), my formula NEVER CONVERGES, but yours converges EVERYWHERE.

So if my formula converges for \(|c| > a\), yours converges for \(1 < | c | <a\). If mine converges for \( |c| >1\), yours never converges (which I can prove). But when mine doesn't converge for all \(|c| > 1\), then yours converges for \(|c|>1\)....

So, considering \(|c| >1\)--where my formula works, yours doesn't, where mine doesn't, yours does... for all \(|c| > 1\). Are they analytic continuations of each other!?!?!?!?!?

I do believe we have a complimentary relationship going on with this shit!

[Leo.W]

Just a thought.
it can be extended to the family of schroder function by
\[\sigma_f(z) = \int_{\mathbb{R}}{h(k)g(f^k(z))\mathrm{d}k}\] (and thus on measurable set in the way that k has some kinda shift-invariance), where \(h\) is any function that fits \(h(k-1)=s\,h(k)\), \(s\) is the multiplier, and \(g\) is any function.

[tommy1729]

Just a thought.
it can be extended to the family of schroder function by
\[\sigma_f(z) = \int_{\mathbb{R}}{h(k)g(f^k(z))\mathrm{d}k}\] (and thus on measurable set in the way that k has some kinda shift-invariance), where \(h\) is any function that fits \(h(k-1)=s\,h(k)\), \(s\) is the multiplier, and \(g\) is any function.

But you cannot do an integral over something with a f^[k](x) dk term because that requires knowing fractional iteration already.

Since you do not have f^[1/2] this seems like a circular reasoning.

So this is a purely theoretical idea and not a computable one .. at first sight.

And also h(k+1) may be c h(k) , but h(k+1/2) is not sqrt© h(k).
And even h(k+1/2)/h(k) is different for most k.
So i see many objections and issues with that.

As for the g(x) part , this might give additional fixpoints and such to limit the region of convergeance or analyticity.

Also I want to warn extra about using g(x) without care :

the following example illustrates why ;

let M(x) = 1/exp(x) + 1/exp(exp(x)) + 1/exp^[3](x) + ...

this M(x) converges for all real x > 0

but it is nowhere convergeant on the complex away from the real line.

Let alone analytic.

If we lower the base of M(x) to smaller then eta but larger than 1 , the 1/ part was not needed and we would have convergeance locally ( x between the fixpoints ) 

Analytic is a desire and intention here.

I understand why ppl like integral representations over sums.
But I think it is psychological :  It WORKED NICELY for standard functions and laplace and fourier stuff.

But for such exotic cases there are many issues and numerical and theoretical problems ... not even mentioning closed forms lacking and analytic issues ...

There are even more issues but I do not want to give an angry impression

regards

tommy1729

[tommy1729]

Just a thought.
it can be extended to the family of schroder function by
\[\sigma_f(z) = \int_{\mathbb{R}}{h(k)g(f^k(z))\mathrm{d}k}\] (and thus on measurable set in the way that k has some kinda shift-invariance), where \(h\) is any function that fits \(h(k-1)=s\,h(k)\), \(s\) is the multiplier, and \(g\) is any function.

But you cannot do an integral over something with a f^[k](x) dk term because that requires knowing fractional iteration already.

Since you do not have f^[1/2] this seems like a circular reasoning.

So this is a purely theoretical idea and not a computable one .. at first sight.

And also h(k+1) may be c h(k) , but h(k+1/2) is not sqrt© h(k).
And even h(k+1/2)/h(k) is different for most k.
So i see many objections and issues with that.

As for the g(x) part , this might give additional fixpoints and such to limit the region of convergeance or analyticity.

Also I want to warn extra about using g(x) without care :

the following example illustrates why ;

let M(x) = 1/exp(x) + 1/exp(exp(x)) + 1/exp^[3](x) + ...

this M(x) converges for all real x > 0

but it is nowhere convergeant on the complex away from the real line.

Let alone analytic.

If we lower the base of M(x) to smaller then eta but larger than 1 , the 1/ part was not needed and we would have convergeance locally ( x between the fixpoints ) 

Analytic is a desire and intention here.

I understand why ppl like integral representations over sums.
But I think it is psychological :  It WORKED NICELY for standard functions and laplace and fourier stuff.

But for such exotic cases there are many issues and numerical and theoretical problems ... not even mentioning closed forms lacking and analytic issues ...

There are even more issues but I do not want to give an angry impression

regards

tommy1729

I thought I have seen that before ...

right !

https://math.stackexchange.com/questions...p-exp-expz

special thanks to mick and sheldon.

regards

tommy1729

[Leo.W]

Just a thought.
it can be extended to the family of schroder function by
\[\sigma_f(z) = \int_{\mathbb{R}}{h(k)g(f^k(z))\mathrm{d}k}\] (and thus on measurable set in the way that k has some kinda shift-invariance), where \(h\) is any function that fits \(h(k-1)=s\,h(k)\), \(s\) is the multiplier, and \(g\) is any function.

But you cannot do an integral over something with a f^[k](x) dk term because that requires knowing fractional iteration already.

Since you do not have f^[1/2] this seems like a circular reasoning.

...

regards

tommy1729

Well tommy, I wasn't focused on the computability, but on the interrelationships between members of schroder functions' family instead and it just generalizes another way of theta-mapping(analytically continued)
You're right tommy, it's piece of * when it comes to computation, most of the time.
* But, I say maybe, we can use non-analytic \(h\) and \(g\) right? in the summation case, we can recover the integral by \(h(z)= s^{-\lfloor z\rfloor}\) and \(g(z)=z\), in this case it theoretically converge in many cases, and what if we use, say, \(g(z)=e^{-\|z\|^2}\)?
Anyway just a thought 

[JmsNxn]

Just a thought.
it can be extended to the family of schroder function by
\[\sigma_f(z) = \int_{\mathbb{R}}{h(k)g(f^k(z))\mathrm{d}k}\] (and thus on measurable set in the way that k has some kinda shift-invariance), where \(h\) is any function that fits \(h(k-1)=s\,h(k)\), \(s\) is the multiplier, and \(g\) is any function.

Hey, Leo

This is entirely correct and what I was hinting at, but I didn't bother to go full Lebesgue!

The correct way to discuss this is:

\[
F_c (z) = \int_{\mathbb{R}}f^{\circ k}(z) \,d\mu ( c)\\
\]

And we would choose a (hopefully quasi Borel) measure \(\mu( c)\); which only depends on \(c\) and \(k\). Which would then give us (hopefully) all possible solutions.

This would say, for all \(\mu( c)\) such that:

\[
F_c(f(z)) = cF_c(z)\\
\]

The value:

\[
\sum_{k=-\infty}^\infty c^k f^{\circ k}(z)\\
\]

Would belong to this class--where there is some \(\mu^*\) which satisfies this (has moles/atoms at \(k \in \mathbb{Z}\)). I believe the official notation would be: \(\mu^*( c) = \sum_{n\in\mathbb{Z}} c^n \delta(k-n)\) (I'm a little out of date on my distribution knowledge, lol )

But, let's play your game, Leo! I was avoiding this discussion. But let's write:

\[
\int_{-\infty}^\infty f^{\circ k}(z) c^k \,dk\\
\]

It turns out this is the "total" holomorphic function to satisfy Tommy's identity. And the only other thing that can is a Fourier transform. What we need is an "indicator" function \(\chi\) that is translationally invariant. This means that \(\chi(k+1) = \chi(k)\). And \(\chi : \mathbb{R} \to \mathbb{C}\). So that:

\[
\int_{-\infty}^\infty f^{\circ k}(z) c^k \chi(k) \,dk \\
\]

Are all solutions. This is assuming that we're considering the ideal case of \(f(z) = \sqrt{2}^z\)--and the standard iteration. So tommy's solution, in and of itself, is the indicator function \(\chi_{\mathbb{Z}}\) on the integers. We can defs choose other indicators though. But they will always be discrete.

This is the super fun part. Imagine we write:

\[
F[\chi_c](z)= \int_{-\infty}^\infty f^{\circ k}(z) c^k \chi_c(k) dk
\]

Now, let's write the standard fourier analysis approach...

\[
F[\chi_c](z) = \sum_{j=-\infty}^\infty b_j( c) F[e^{-2\pi i j k}](z)\\
\]

Which happens because every element of Tommy's space of solutions, can be found as the integral:

\[
F[\chi_c](z)= \int_{-\infty}^\infty f^{\circ k}(z) c^k \chi_c(k) dk
\]

And each:

\[
\chi_c(k) = \sum_{j=-\infty}^\infty b_j(c ) e^{-2 \pi i j k}\\
\]

BUT!!!!!!!!!!!!!!

This only happens in a distributional sense. In a technical sense, the value \(\mu^*(c )\) cannot be represented as this. When I write:

\[
F[\chi_{\mathbb{Z}}](z) = \int_{-\infty}^\infty f^{\circ k}(z) c^k \chi_{\mathbb{Z}}(k) dk = \sum_{k=-\infty}^\infty f^{\circ k}(z)c^k\\
\]

We do not mean that the naive numerical approximation methods work. What we mean is that this converges in a distributional sense. This means we have to introduce an \(L^2\) space. In this case, we can take something standard, like the standard Fourier Analysis space \(L^2(\mathbb{R})\). And from this, we have the inner product:

\[
(f,g) = \int_{-\infty}^\infty f(k)\overline{g(k)}\,dk\\
\]

And from the extension of the Hilbert space (using Von Neumann magic), we can define:

\[
F[\chi_c](z) = (f^{\circ k}(z) c^k, \chi_c(k))\\
\]

And as a distribution; we can Fourier decompose \(\chi_c\), even though it may not be obvious....

Tommy's \(\chi_{\mathbb{Z}}\) would be the same thing as a "square wave". It can be approximated with things underneath the integral....

[tommy1729]

Just a thought.
it can be extended to the family of schroder function by
\[\sigma_f(z) = \int_{\mathbb{R}}{h(k)g(f^k(z))\mathrm{d}k}\] (and thus on measurable set in the way that k has some kinda shift-invariance), where \(h\) is any function that fits \(h(k-1)=s\,h(k)\), \(s\) is the multiplier, and \(g\) is any function.

Hey, Leo

This is entirely correct and what I was hinting at, but I didn't bother to go full Lebesgue!

The correct way to discuss this is:

\[
F_c (z) = \int_{\mathbb{R}}f^{\circ k}(z) \,d\mu ( c)\\
\]

And we would choose a (hopefully quasi Borel) measure \(\mu( c)\); which only depends on \(c\) and \(k\). Which would then give us (hopefully) all possible solutions.

This would say, for all \(\mu( c)\) such that:

\[
F_c(f(z)) = cF_c(z)\\
\]

The value:

\[
\sum_{k=-\infty}^\infty c^k f^{\circ k}(z)\\
\]

Would belong to this class--where there is some \(\mu^*\) which satisfies this (has moles/atoms at \(k \in \mathbb{Z}\)). I believe the official notation would be: \(\mu^*( c) = \sum_{n\in\mathbb{Z}} c^n \delta(k-n)\) (I'm a little out of date on my distribution knowledge, lol )

But, let's play your game, Leo! I was avoiding this discussion. But let's write:

\[
\int_{-\infty}^\infty f^{\circ k}(z) c^k \,dk\\
\]

It turns out this is the "total" holomorphic function to satisfy Tommy's identity. And the only other thing that can is a Fourier transform. What we need is an "indicator" function \(\chi\) that is translationally invariant. This means that \(\chi(k+1) = \chi(k)\). And \(\chi : \mathbb{R} \to \mathbb{C}\). So that:

\[
\int_{-\infty}^\infty f^{\circ k}(z) c^k \chi(k) \,dk \\
\]

Are all solutions. This is assuming that we're considering the ideal case of \(f(z) = \sqrt{2}^z\)--and the standard iteration. So tommy's solution, in and of itself, is the indicator function \(\chi_{\mathbb{Z}}\) on the integers. We can defs choose other indicators though. But they will always be discrete.

This is the super fun part. Imagine we write:

\[
F[\chi_c](z)= \int_{-\infty}^\infty f^{\circ k}(z) c^k \chi_c(k) dk
\]

Now, let's write the standard fourier analysis approach...

\[
F[\chi_c](z) = \sum_{j=-\infty}^\infty b_j( c) F[e^{-2\pi i j k}](z)\\
\]

Which happens because every element of Tommy's space of solutions, can be found as the integral:

\[
F[\chi_c](z)= \int_{-\infty}^\infty f^{\circ k}(z) c^k \chi_c(k) dk
\]

And each:

\[
\chi_c(k) = \sum_{j=-\infty}^\infty b_j(c ) e^{-2 \pi i j k}\\
\]

BUT!!!!!!!!!!!!!!

This only happens in a distributional sense. In a technical sense, the value \(\mu^*(c )\) cannot be represented as this. When I write:

\[
F[\chi_{\mathbb{Z}}](z) = \int_{-\infty}^\infty f^{\circ k}(z) c^k \chi_{\mathbb{Z}}(k) dk = \sum_{k=-\infty}^\infty f^{\circ k}(z)c^k\\
\]

We do not mean that the naive numerical approximation methods work. What we mean is that this converges in a distributional sense. This means we have to introduce an \(L^2\) space. In this case, we can take something standard, like the standard Fourier Analysis space \(L^2(\mathbb{R})\). And from this, we have the inner product:

\[
(f,g) = \int_{-\infty}^\infty f(k)\overline{g(k)}\,dk\\
\]

And from the extension of the Hilbert space (using Von Neumann magic), we can define:

\[
F[\chi_c](z) = (f^{\circ k}(z) c^k, \chi_c(k))\\
\]

And as a distribution; we can Fourier decompose \(\chi_c\), even though it may not be obvious....

Tommy's \(\chi_{\mathbb{Z}}\) would be the same thing as a "square wave". It can be approximated with things underneath the integral....

correct.

[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.

---

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 \(0 < \Im(s) < \frac{2\pi}{\lambda}\).

---

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...

---

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!

---

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!

---

Deepest Regards, James

---

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

---

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!

[JmsNxn]

Let's take \(z \in U\), such that \(f^{\circ k}(z) \to 2\) and \(f^{\circ -k}(z) \to 4\), as \(k\to\infty\)--where \(f(z) = \sqrt{2}^z\). The set \(U\) is simply connected domain, and subject to all our "double fixed point talk". Then taking the regular iteration for \(f:U \to U\). This means that \(f^{\circ s} : U \to U\) for \(\Re s > 0\).

Then:

\[
\widehat{f^{\circ s}}(\xi,z) = \int_{-\infty+it}^{\infty+it} \left( f^{\circ s}(z) -2\right) e^{-2\pi i \xi s}\,ds\\
\]

Applications of \(f\) become applications of a wave multiplier \(e^{-2\pi i \xi}\). Which is the equation:

\[
\widehat{f^{\circ s}}(\xi,f^{\circ q}(z)) = e^{2\pi i \xi q} \widehat{f^{\circ s}}(\xi,z)
\]

We can prove this in a very strong sense. The heavy lifting is handled by Henryk, and Dmitri. The paper they wrote on the Four super exponentials of \(\sqrt{2}^z\) proves everything we need. I'd put a lot of the credit to Henryk--because I had written integrals similar before. But Henryk has paved a way to absolutely, rigorously, prove it.