[MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 03/11/2023, 11:27 PM https://math.stackexchange.com/questions...expz-expfz f(f(z)) = z , f(exp(z)) = exp(f(z)) regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 03/14/2023, 11:23 PM bell polynomials might help. regards tommy1729 Daniel Fellow Posts: 248 Threads: 80 Joined: Aug 2007 03/15/2023, 12:53 AM Trivially $$f(z)=z.$$ Daniel JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 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 tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 03/17/2023, 12:48 AM (03/16/2023, 03:56 AM)JmsNxn Wrote: 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 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 tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 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. JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 03/18/2023, 03:37 AM (This post was last modified: 03/18/2023, 03:48 AM by JmsNxn.) 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 unequivalent--which 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). Daniel Fellow Posts: 248 Threads: 80 Joined: Aug 2007 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 JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 03/18/2023, 05:54 AM (03/18/2023, 05:16 AM)Daniel Wrote: 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. 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! « Next Oldest | Next Newest »

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