Flow and convergence

[Daniel]

The iteration of holomorphic functions are holomorphic functions.
If a holomorphic function is described by a flow \[f^s(f^t(z))=f^{s+t}(z)\] then it converges where \[s,t\in\mathbb{N}\]. Also since \[f^s\] converges for \[s\in\mathbb{N}\] and the iteration of holomorphic functions are holomorphic functions, then it converges where \[s,t\in\mathbb{Q}\].
Since rational numbers can be taken arbitrarily close to each other while being convergent, an irrational number can only be divergent if the first derivative goes to infinity. 
Since a first derivative of a holomorphic function can't go to infinity except at infinity,
\[f^s(f^t(z))=f^{s+t}(z)\implies s,t \in \mathbb{R}\] converges.

[JmsNxn]

Hey, Daniel; you have described flows:

\[
\frac{d}{dt} y(t) = F(y(t))\\
\]

Where:

\[
F(z) = \frac{d}{dx}\Big{|}_{x=0} f^{\circ x}(z)\\
\]

.....

Now let's describe the path \(y(t)\).  Let's call \(y(t) = f^{\circ t}(z)\). This is written as a first order differential equation, which can be translated to a flow equation:

\[
y'(t) = F(y(t))\\
\]

If we trace any path; we hit all the solutions, and hit discontinuities if they are natural....

Any flow equation like this allows us to write \(f^{\circ s}(f^{\circ t}) = f^{\circ s +t}\). But we can start from my perspective, then find this equation naturally.

This is one of the motivations of my "differential bullet product" notation.


....

You can also just write this as matrices multiplying.........

I write this as:

\[
f^{\circ t}(z) = \int_0^t F(z)\,ds\bullet z\\
\]

Which is just an advanced integral of the matrix interpretation....

[tommy1729]

hello guys

You seem to be missing the fact that this IS EQUAL TO THE SEMI-GROUP ISOM I have been talking about for months.

This thing is a uniqueness criterion !

Also real-differentiable may not be equal to complex differentiable.

And in older posts I already called something flow theory which somewhat relates.

But yeah , analytic over the rationals implies real-analytic over the reals for smooth functions.

If you can show the function as smooth over the complex and analytic over the gaussian or eisenstein rationals then it is analytic.

This is just good old basic analysis.



The 2sinh has this property by the fact of its fixpoint at 0.

the compositions of iterations of 2sinh with ln and exp are just conjugate compositions , so the property carries over from the 2sinh super to the super of the 2sinh method.
It is known to be C^oo so it is smooth over the rationals.

Question is if it is analytic.

Im unaware of nontrivial cases where semi-group iso that is not analytic.
I posted conjectures about these in the past. 


regards

tommy1729

[tommy1729]

the matrix multiplication has issues if you mean carleman and such.

This follows from the many sqrt a matrix can have.

if the eigenvalues are nonreal the n th root of the matrix can have n solutions PER eigenvalue.

So the n th root of a m*m matrix can have up to 

n^m

SO 

a matrix of size 10 can have

2^10 sqrt roots.

and that does not consider the one-periodic function.

So unless you define a way to get the correct roots ( such as closest to the + reals )
and that works out nice , even with increasing matrix size

then and only then  may you have a good definition.



It has been suggested to use 

A^(1/n) = exp( 1/n ln(A) )



this way maybe you have the semi group iso , IF the ln and exp do not give issues.
( again by conjugate composition )

But this seems only consistant when locally around a fixpoint :

so basically equivalent to the fixpoint expansion like koenings function ( assuming f ' (fix ) is ok )



regards

tommy1729