Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\)
#13
Hi, I hate to waste anyone time but in some place I'm really puzzled by really obvious things...  my math is not that strong prolly. I have many obstacles in getting this.

1) Let's see if I get this right when you say "we know that \(e^{i\theta q} = e^{2 \pi i p - qt} \in (0,1)\)...
Let \(\theta \in \mathbb H^*=2\pi\mathbb Q+\mathbb R^+ i\), then there exists a minimal \(q\in\mathbb N\) s.t. we have the form \(\theta q=2\pi p+itq \in 2\pi\mathbb N+\mathbb R^+ i \). Clearly \(q_\theta\) depends on theta... we could call it the d-part of \(\theta\), call it \(q={\sf d}(\theta)\). We  have \({\sf d}:\mathbb H^*\to\mathbb N\)

note for myself: \(\mathbb H^*\) is closed under addition, but since it has not zero in it is not a monoid, just a semigroup...

Now you say that \((e^{i\theta})^{{\sf d}(\theta)}\in (0,1)\) but its not obvious... if I compute it as \(e^{i(\theta {\sf d}(\theta) )}=e^{i(2\pi p+it{\sf d}(\theta))}=e^{i2\pi p}/ e^{t{\sf d}(\theta)} \)

then, maybe, since the first factor has exponent a multiple of 2pi we get \[ e^{i(\theta {\sf d}(\theta) )}=1/ e^{t{\sf d}(\theta)}\]

2) Then something seems off with the formula there. Imho. Set \(\ell(\theta)=\frac{2\pi}{\theta {\sf d}(\theta) }=\frac{2\pi}{2\pi p +it{\sf d}(\theta) }\) and consider the function \(E(s):=(e^{i\theta  {\sf d}(\theta) })^s\), clearly it is an exponential with \(E(1)=e^{i\theta  {\sf d}(\theta) }\) as you are setting up and we get some identities

\[E(-s)=e^{i\theta  {\sf d}(\theta)(-s)}=(e^{t{\sf d}(\theta)})^{s}\]
\[E(s+t)=E(s)\cdot E(t)\]

And we get a group homomorphism.... And then the periodicity, maybe it is just a typo in your formula?
Periodicity is about finding elements in the kernel of the homomorphism. If the kernel is nontrivial it contains non-zero \(l\in {\rm ker}E\) s.t. \(E(k)=1\). This implies that \(E(s+l)=E(s)\). The kernel of a group morphism must be itself a subgroup of the domain group... it measures the injectivity in a sense. Also the kernel is closed under scalar multiplication by integers. So we have many periods: if \(l\) is in the kernel, call it a period, then \(E(al)=E(l)^a=1\) this means that the kernel is a integral vector space... has is a generating set/basis?

\[E(s+\ell(\theta))=e^{i\theta  {\sf d}(\theta)(s+\ell(\theta))}=E(s)\cdot e^{i\theta  {\sf d}(\theta)\frac{2\pi}{\theta {\sf d}(\theta) }}=E(s)\]

.... up unitl this point i'm almost there but I feel weak. Skimming ahead I see I'll need some time and effort to put all the pieces together... so just more questions before I give my try:

3) I dont get the definition of the set \(\mathcal N\), what is \(\delta\)? Is it dependent on the choice of the arguments...

4) at the end I kinda get the big picture... we are working on a subset generated by rational multiples in order to approximate the function on a bigger, complete, set. But I still miss why we are doing this. To obtain a  fractional iteration of... the multiplication? You are laying this out as a toy model so that you can apply the whole machinery on different functions?

All of this seems a very long story. Can you also streamline the structure of the construction... the big picture.
For example, I don't remember anymore how this is related with Euler identity.

Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)

\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)
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Messages In This Thread
RE: Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\) - by MphLee - 01/06/2023, 07:05 PM

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