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+--- Thread: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) (/showthread.php?tid=1726)
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[MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - tommy1729 - 03/11/2023
https://math.stackexchange.com/questions/4656525/ffz-z-f-expz-expfz
f(f(z)) = z , f(exp(z)) = exp(f(z))
regards
tommy1729
RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - tommy1729 - 03/14/2023
bell polynomials might help.
regards
tommy1729
RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - Daniel - 03/15/2023
Trivially \[f(z)=z.\]
RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - JmsNxn - 03/16/2023
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
RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - tommy1729 - 03/17/2023
(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
RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - tommy1729 - 03/17/2023
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.
RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - JmsNxn - 03/18/2023
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).
RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - Daniel - 03/18/2023
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,
\[0^0=1, 0^1=0\]
![[Image: StandardGrid.gif]](https://tetration.org/original/Fractals/Atlas/StandardGrid.gif)
![[Image: MOONJ002.gif]](https://tetration.org/original/Fractals/Atlas/Pseudocircle/MOONJ002.gif)
\[f(exp(z)) = exp(f(z)) \implies f(z) = \, ^{z+n}e\] 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.
RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - JmsNxn - 03/18/2023
(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!
RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - tommy1729 - 04/03/2023
(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.
Im sorry, I do not get what you are doing ?
f(exp(z)) = exp(f(z)) does suggest that f being an iteration of tetration base e is a solution.
However not being base e and having f(f(z)) suggest not being an iteration of exp(x).
But it does not show that f(x) does not exist ?
By analogue :
g(h(h(x))) = g(h(x)) = g(x)
does not imply h(x) is an iteration of g.
for instance
h(x) = 1/x
g(x) = (x^4 + 3 x^2 + 1)/(x^3 + x)
Although It is a fact that g(h(x)) =/= h(g(x)).
So the analogue is not perfect.
***
I see you could say that
e F(x) = F(exp(x))
Implies that
- F(x) = F(f(x))
where f(f(x)) = x
exp(f(x)) = f(exp(x))
And therefore
sexp(x + pi i) = f( sexp(x) )
for a 2pi i periodic sexp at least in a strip containing Real +/- 2pi i.
Im not talking about global solutions , valid everywhere and analytic everywhere.
But local solutions valid somewhere and locally analytic.
***
f(exp(x+2pi i)) = f(exp(x)) = exp(f(x))
so f(x) = f(x + 2pi i) + 2 pi i K.
or not valid or analytic in such a range.
***
f(exp(x)) = exp(f(x))
implies LOCALLY
ln( f(exp(x)) ) + 2 pi i L = f(x)
All of these observations are based on the invariant of exp(x).
***
Another idea I had is this one :
We know exp(x) = lim (1+ x/n)^n
And every solution f_n(x) to
f_n( (1+x/(n+3))^(n+3) ) = (1 + f_n(x)/(n+2))^(n+2)
exists for every finite positive integer n.
If f_n(x) converges for increasing n, and has a non-zero radius and validity for some x , the limit is a solution.
Another idea is truncated taylors
f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_m x^m
f(f(x)) = x mod x^n
f(exp(x)) = exp(f(x)) mod x^n
for m/2 < n < m and increasing m.
Assuming the error terms can be neglected and the equation converges.
And perhaps this MOD idea is relevant too ?
https://math.eretrandre.org/tetrationforum/showthread.php?tid=1735
Now you may be right and I think sheldon once proved a similar statement.
Maybe I forgot.
Sorry for the dumb question maybe
regards
tommy1729