Borel summation, Mellin Transforms, Parabolic iteration

[bo198214]

Oh yes, sorry. I misplaced my terms. What I meant is still the same as you though, I just thought the multiplicity = valit. Either way I meant that:

\[
f(z) = z(1+az^n + o(z^n))\\
\]

Yes, no biggie. Just wanted us to be clear about the terms.

Don't you need more terms \(\frac{1}{u^k}\) and also having coefficients, like:

\[ \alpha(u) = \left(\sum_{k=1}^{\ell} \frac{c_k}{u^k}\right) - A \log(u) + \sum_{k=0}^\infty b_k u^k \]

I believe your mistake is thinking we have to conjugate the Abel function--which we don't do. We conjugate the function the abel function is associated with--within the Abel function. By which, I mean:

No, actually I didnt care about the conjugations, I were just relying on

We have an abel function on half plane, which is really just the abel function of our function \(f\), but fixated in a petal, and near \(0\).

and the Abel function of \(f\) looks like I described above ... but with all those conjugations ...
I mean you first conjugate with \(\varphi(z) = \frac{\mathbf{v}}{a\sqrt[\ell]{z}}\) but then you conjugate with \(1/z\) which would be imho the conjugation \(\frac{a\sqrt[\ell]{z}}{\mathbf{v}}\) ? Something does not yet feel right about the Abel formulas (apart from that they don't converge of course.)