03/11/2023, 11:27 PM
https://math.stackexchange.com/questions...expzexpfz
f(f(z)) = z , f(exp(z)) = exp(f(z))
regards
tommy1729
f(f(z)) = z , f(exp(z)) = exp(f(z))
regards
tommy1729
[MSE] f(f(z)) = z , f(exp(z)) = exp(f(z))

03/11/2023, 11:27 PM
https://math.stackexchange.com/questions...expzexpfz
f(f(z)) = z , f(exp(z)) = exp(f(z)) regards tommy1729
03/14/2023, 11:23 PM
bell polynomials might help.
regards tommy1729
03/16/2023, 03:56 AM
This reduces into a problem with riemann mappings.
We can write tommy's equation as: $$ f(g(f^{1}(z)) = g(z)\\ $$ These equations are only solvable if \(g : D \to D\). Then we need that \(f : D \to D\). The only domain \(g = \exp\) takes to itself is \(\mathbb{C}\). Trying to create conjugations on the entire complex plane is a null effort. There exists no conjugations in this case. The cases where you can create conjugations on all of \(\mathbb{C}\), always require a smaller domain \(D\) that satisfies this. If \(D\) were simply connected and biholomorphic to the unit disk; then we can write the solutions for: $$ f(g(f^{1}(z))) = h(z)\\ $$ Very simply for all \(f,g,h\). But as \(\exp(z)\) only fixes the complex plane, this isn't possible; unless you introduce a lot of branching talk. So, Daniel is right in saying that \(z \mapsto z\) is the trivial solution, but it's also the only solution on \(\mathbb{C}\). To get a different solution, we can take something like Kneser's: $$ f(z) = \text{sexp}_K(\text{slog}_K(z) + 1/2)\\ $$ But this is not holomorphic on \(\mathbb{C}\). And we get a chicken and the egg situation... Do we solve tetration, or do we solve this conjugation first? I'm not sure how helpful it is to look at this first; rather than looking at iterates which appear more natural; but I could be wrong... Regards, James
03/17/2023, 12:48 AM
(03/16/2023, 03:56 AM)JmsNxn Wrote: This reduces into a problem with riemann mappings. Minor silly comment : exp maps C to C \ {0} But yeah that may be an issue. also quote : $$ f(z) = \text{sexp}_K(\text{slog}_K(z) + 1/2)\\ $$ end quote I think you meant $$ f(z) = \text{sexp}_K(\text{slog}_K(z) + (1)^{1/2})\\ $$ $$ f(z) = \text{sexp}_K(\text{slog}_K(z) + i)\\ $$ or maybe $$ f(z) = \text{sexp}_K(\text{slog}_K(z) + 1/2 \pi i)\\ $$ so complex ? right ? afterall f(f(z)) = z and the periodicity in the complex direction would make f and exp commute potentially. Or maybe im missing something. ( my mistake ) regards tommy1729
03/17/2023, 12:53 AM
Mick edited his question with a carleman matrix idea.
both on MSE and MO. regards tommy1729 ps : this actually shows ( i think ) that the carleman matrix method does not agree with the infinite composition methods ( early 2020 ?) having 2pi i periodicity.
I apologize, I forgot to add an additional element of algebra. I forgot that you are requiring \(f(f(z)) = z\). For that you have to go a little tricky. I'm going to assume that \(\Im(y) \neq 0\) and that \(\overline{y} =  y\).
I meant that: $$ f(z) = \text{sexp}_K\left( \text{slog}_K(z)+ y\right)\\ $$ Is holomorphic on \(\mathbb{C} / S\), where \(S\) is a set of singularities and branches. But: $$ f(\exp(z)) = \exp(f(z))\\ $$ To get the function you want we have to be way more clever. Sorry, my eyes must've glazed over before. I think a viable solution will require using the use of the conjugation function; So if I take: $$ F(z) = f(\overline{f^{1}(\overline{z})}) \neq z\\ $$ This is a holomorphic function because Sexp is conjugate symmetric. But then this is also satisfying: $$ F(F(z)) = f(\overline{f^{1}(\overline{f(\overline{f^{1}(\overline{z})}))}}) =z\\ $$ Unfortunately, I think this is still just Daniel's trivial solution \(z\), just repackaged in a fancy way; I'm sure you can find more like this. Also, I may have fucked up some switch ups, lmao. This should be \(z\)... I think, lmao. I apologize; don't think I have much to add to this. Well considering the infinite composition method/beta method is nowhere holomorphic for \(b = e\); I'm not surprised it's nowhere equivalent to the Carlemann method. Even when you choose arbitrary solutions; the beta method is unequivalentwhich was kind of the point. Though it can be holomorphic, and if it was holomorphic it was unique; Carlemann always chooses the Schroder or Kneser approach. Which is regular iteration or crescent iteration. I apologize, long day. Sorry If I'm not making any sense. I do think the switch from upper half plane to lower half plane is the trick to this problem (when using Kneser).
03/18/2023, 05:16 AM
Tetrational Geometry
The equation $$f(exp(z)) = exp(f(z))$$ tells us we want to be looking at tetration. Now add the constraint $$f(f(z))=z$$ in the context of tetration. This means we are looking at period 2 tetration. The orange disk in the following image and the red disk in the image after is the area where tetration displays period 2 behavior. See Pseudocircle The number 0 lies in the disk and is period 2 under tetration, $$0^0=1, 0^1=0$$ $$f(exp(z)) = exp(f(z)) \implies f(z) = \, ^{z+n}e$$ which is inconsistent with the complex base being within the pseudocircle. The question could be generalized to $$f(a^z) = a^{f(z)} \textrm{ and } f(f(z))=z \textrm{ where } a \textrm{ is in the pseudocircle.}$$. Since $$a \ne e$$ the question has no solution beyond the trivial solution.
Daniel
03/18/2023, 05:54 AM
(03/18/2023, 05:16 AM)Daniel Wrote: Tetrational Geometry Fucking beautiful, Daniel. That was my suspicion; your solution is the only solution. I was trying to fiddle with shit; but I was pretty sure it was still \(z\). Nice! 
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