(08/21/2022, 08:54 AM)bo198214 Wrote:(08/21/2022, 01:18 AM)JmsNxn Wrote: Quick question, I'm a little confused here.We are looking at the growth of the coefficients of the logit, which shows quite the similar behaviour as the coefficients of the half root,
Is this still guessing the asymptotics of a "half root" at a parabolic fixed point?
Or is it something different, (sorry just a tad confused).
but speaks so to says about all *iterates* not only the half root. I didn't test it but I am super sure the 3rd root and 4th root, etc show the same behaviour - so the idea was to make it independent of the concrete iterative root, just dependent on the original function.
The logit is a good candidate for this - because the logit is analytic at the fixed point if and only if all regular iterates are analytic at the fixed point.
So I gave more examples for arbitrarily chosen parabolic functions where the logit shows a similar coefficient growth pattern as the logit coefficients of \(e^x-1\), where the concrete growth behaviour depends on the \(x^2\)-coefficient of the function.
But then I also gave a counter example of a parabolic function where the logit is analytic (and hence all the regular iterates), i.e. the formal powerseries converges, i.e. the coefficients don't show that growth behaviour and don't need divergent summation.
And in the last post I just looked at the logit and it's relations for this counter-example from a non-powerseries pov and how you reconstruct the Abel function from the logit.
(08/21/2022, 01:18 AM)JmsNxn Wrote: If this is happening elsewhere though; maybe Borel summation would be a valuable method of approaching fractional iteration?
By which we could get similar Euler expressions (Like how Euler analytically defines \(\sum_k (-1)^kk! z^k\)) of half iterates (and arbitrary iterates) using some kind of modified Laplace transform. All we would need is a bound like \(j_k = O(c^kk!)\).
Actually I wonder why Gottfried didnt post any results about the divergent summations he tried.
To get the left-side and the right-side iterates maybe one needs to apply two different divergent-summations
Ok I see, I was confused.
I'd just like to add that this use of logit, is something Mphlee and I wanted to unite with Differential equations, and as the generator of a semi group.
Every differential equation:
\[
y '= f(y)\\
\]
Induces a Semi group. And every semigroup \(y(t,z)\) induces a differential equation:
\[
\frac{d}{dt} y(t,z) = f(y(t,z))\\
\]
By which I spent a significant portion of one of my reports discussing taking fourier transforms on these operations. By which, every semi-group in this form can be written:
\[
\begin{align}
y(t,z) &= g(t+g^{-1}(z))\\
f(z) &= \frac{d}{dt}\Big{|}_{t=0} y(t,z)\\
\end{align}
\]
Where we can take fourier transforms across these objects, by writing:
\[
\mathcal{F}\{y(p(t),z)\}(\xi,z) = g( \int_{-\infty}^\infty p(t)e^{2 \pi i t\xi}\,dt + g^{-1}(z))\\
\]
Where this is compositionally linear, in the sense that:
\[
\begin{align}
\mathcal{F}\{y(p(t),z)\}(\xi,z) &= h_1(\xi,z)\\
\mathcal{F}\{y(q(t),z)\}(\xi,z) &= h_2(\xi,z)\\
h_1(\xi,h_2(\xi,z)) &= h_1 \bullet h_2 \bullet z = \mathcal{F}\{y(p(t) + q(t),z)\}(\xi,z)\\
\end{align}
\]
This operation is invertible in many circumstances. For example, the fourier transfrom on the semigroup induced by \(f(z) = z^2\), is given as:
\[
\frac{1}{\frac{1}{z}-\int_{-\infty}^\infty p(t)e^{2 \pi i t\xi}\,dt}\\
\]
Where additionally, We have the convenient identity:
\[
\frac{d}{dt} y(p(t),z) = \frac{d}{dt}\frac{1}{\frac{1}{z}-p(t)} = p'(t) f(y(p(t),z)) = p'(t)y(p(t),z)^2\\
\]
And we have a relationship between Separable cases and semi-group cases. This is largely aside to your discussion, but MphLee and I wanted to rigorize a lot of this discussion, as it was mostly only in theory form. And I had no obvious way of solving/numerically evaluating for this Fourier transform in a general scenario.
And this is expressible through "compositional integrals" (which is kind of just a fancy way of doing Euler's method but in the complex plane, and only for holomorphic functions).
I'm very interested by this, especially if we can tie it into mellin transforms; and Borel sum the logit.
Anyway, didn't mean to detract from the discussion at hand. This seems very promising though!!! Lol, this makes me think I might be able to find a practical expression for this "semi-group action fourier transform"--rather than the algebraic nonsense I currently have, lol.
