Borel summation, Mellin Transforms, Parabolic iteration
#5
I might be making a mistake somewhere. I know the mellin transforms don't necessarily converge; but they necessarily produce a mellin transformed function. This sort of goes "distributional theory", but not totally.


We get that--\(f(0) = 0\), and \(f'(0) = 1\), valit 1. There by we can rewrite this \(F(w) = 1/f(1/w)\). Now from here we have a half plane \(H_R\) for \(\Re(w) > R\), where \(F(w) \sim w + 1 + o(w)\), whereby \(F\) is injective on \(H_R\). Now let's do the conjugation \(G(w) = F(w+R) - R\). Now, \(G\) is holomorphic on a right half plane \(\Re(w) > 0\). What I had produced as my conclusions, were based on this.

From here, we can make another switch \(G(u) = G(1/w)\), which fully produces our expansion at infinity if \(B(G(u)) = B(u) + 1\), for a function \(B\). And \(B\) is the same thing, it is an LFT \(\ell\) applied to \(\alpha(f)\). It looks something like \(\alpha(\ell(u)) = B(u)\). Where \(\ell(G) = f(\ell)\).

Can I ask what you are trying to numerically evaluate, so I can chime in? I don't know how you're approaching it. This would be exceedingly difficult to program in. I can't think of any obvious manner of evaluating this efficiently and effectively. You have to find \(R\), you have to perform conjugations, and require taylor data at every step.
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RE: Borel summation, Mellin Transforms, Parabolic iteration - by JmsNxn - 09/10/2022, 08:00 AM

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