(01/06/2023, 01:28 AM)JmsNxn Wrote: 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