1.)
I, too, have looked at trying to identify the theta mapping in other ways, using "some kind of" fourier transform. This would be done with the Mellin transform, or the Laplace transform. It turns out similar to the following.
If \(F(z)\) is a super function, that satisfies \(f(F(z)) = F(z+1)\). Then assume additionally that \(F\) is holomorphic in a half plane, WLOG we can set the half plane as \(\Re(z) > -\delta\). This would equate to real valued multipliers, but complex valued multipliers can be found through a change of variables. Then we can show that:
\[
\frac{1}{2\pi i} \int_{k-i\infty}^{k+i\infty} F(-z)\Gamma(z)x^{-z}\,dz\\ = \sum_{n=0}^\infty f^{\circ n}(F(0))\frac{x^n}{n!}\\
\]
For \(0 < k < \delta\). You'll note this is a fourier transform in disguise.
Well, it just so turns out. The only function this operation converges on is when \(\theta = \text{Constant}\). I used this as a uniqueness condition in much of my earlier work. I've never quite phrased it as this, but this is a consequence of Ramanujan's master theorem, and the fact \(f^{\circ n}(F(0)) = F(n)\) regardless of \(\theta\). To go against this, would require that we are not holomorphic in a half plane.
The other kind of "fourier transform" I thought, after this, was less restrictive--let's just use Laplace. If we assume that \(F\) is bounded, and about an attracting fixed point \(\lim_{t \to \infty} F(t) = A\). Then:
\[
\int_0^\infty e^{-st}F(t)\,dz = LF(s)\\
\]
I don't remember what I worked out here, but I found a way to describe \(LF\)'s dependence on \(\theta\).
2.)
Also, when it comes to first order difference equations, I've done a lot of work on it. Which probably culminates in a paper I wrote back in 2019. I can link it if you're interested. It's primarily focused on solving the equation:
\[
\Delta f(s) = e^{sf(s)}\\
\]
But it allows for solutions of lots of whacky first order difference equations (and even higher order ones). I just chose that equation as a case study to describe how it's done more generally. Though it does take a deep deep dive into infinite compositions and how they work. And that's largely the heavy lifting.
Could you further elaborate how a First Order Difference Equation can help solve a super function problem? Because I always got hung up on how these things seemed irreconcilable. The closest I got was the Beta method. Which solves the first order difference equation:
\[
\Delta \beta = \frac{e^{\beta(s)}}{1+e^{-s}} - \beta(s)\\
\]
Which is slightly different than your form. I'm very interested in this. Additionally it was shown that in general, these solutions are not holomorphic (barring Schroder solutions). So it's no help for base \(e\) for example (at least real valued ones).
I, too, have looked at trying to identify the theta mapping in other ways, using "some kind of" fourier transform. This would be done with the Mellin transform, or the Laplace transform. It turns out similar to the following.
If \(F(z)\) is a super function, that satisfies \(f(F(z)) = F(z+1)\). Then assume additionally that \(F\) is holomorphic in a half plane, WLOG we can set the half plane as \(\Re(z) > -\delta\). This would equate to real valued multipliers, but complex valued multipliers can be found through a change of variables. Then we can show that:
\[
\frac{1}{2\pi i} \int_{k-i\infty}^{k+i\infty} F(-z)\Gamma(z)x^{-z}\,dz\\ = \sum_{n=0}^\infty f^{\circ n}(F(0))\frac{x^n}{n!}\\
\]
For \(0 < k < \delta\). You'll note this is a fourier transform in disguise.
Well, it just so turns out. The only function this operation converges on is when \(\theta = \text{Constant}\). I used this as a uniqueness condition in much of my earlier work. I've never quite phrased it as this, but this is a consequence of Ramanujan's master theorem, and the fact \(f^{\circ n}(F(0)) = F(n)\) regardless of \(\theta\). To go against this, would require that we are not holomorphic in a half plane.
The other kind of "fourier transform" I thought, after this, was less restrictive--let's just use Laplace. If we assume that \(F\) is bounded, and about an attracting fixed point \(\lim_{t \to \infty} F(t) = A\). Then:
\[
\int_0^\infty e^{-st}F(t)\,dz = LF(s)\\
\]
I don't remember what I worked out here, but I found a way to describe \(LF\)'s dependence on \(\theta\).
2.)
Also, when it comes to first order difference equations, I've done a lot of work on it. Which probably culminates in a paper I wrote back in 2019. I can link it if you're interested. It's primarily focused on solving the equation:
\[
\Delta f(s) = e^{sf(s)}\\
\]
But it allows for solutions of lots of whacky first order difference equations (and even higher order ones). I just chose that equation as a case study to describe how it's done more generally. Though it does take a deep deep dive into infinite compositions and how they work. And that's largely the heavy lifting.
Could you further elaborate how a First Order Difference Equation can help solve a super function problem? Because I always got hung up on how these things seemed irreconcilable. The closest I got was the Beta method. Which solves the first order difference equation:
\[
\Delta \beta = \frac{e^{\beta(s)}}{1+e^{-s}} - \beta(s)\\
\]
Which is slightly different than your form. I'm very interested in this. Additionally it was shown that in general, these solutions are not holomorphic (barring Schroder solutions). So it's no help for base \(e\) for example (at least real valued ones).