[MSE] f(f(z)) = z , f(exp(z)) = exp(f(z))

[tommy1729]

https://math.stackexchange.com/questions...expz-expfz

f(f(z)) = z , f(exp(z)) = exp(f(z))



regards

tommy1729

[tommy1729]

bell polynomials might help.

regards

tommy1729

[Daniel]

Trivially $$f(z)=z.$$

[JmsNxn]

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]

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]

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]

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]

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.

[JmsNxn]

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!