(08/27/2022, 07:41 AM)bo198214 Wrote: Great dig, James!
Thanks, Bo, I had to read a lot of shit, and sift through totally unrelated ideas to get here.
Quote:So is there something like a radius of convergence for Borel-summations? How far from the fixed point would it converge/give correct values?
And then I think it would give the correct solution for each petal, right James?
EDIT: It would be really interesting where the break between the petals is when one does the Borel summation ...
I don't think we can find a uniform bound \(\rho > 0\) such that \(|\mathcal{B}g(z)| < \rho\). Where \(g\) is allowed to wander. So, the best I can say is that; yes, you've answered the question--by that comment. The radius of convergence depends on the bounds of other variables. There's still a lot of algebra/math to unpack though. But we are performing all these limits for sectors near zero.
And yes, Borel summing on the julia set; the break in the petals; will be very fucking interesting
And we should absolutely be able to write this using integral transforms. We just have to map the petal to the half plane; as you so poignantly remarked. Any \(\mathcal{P}\) is mappable to \(\mathbb{C}_{\Re(z) < 0}\)--and can be done biholomorphically. Anywhere you see Borel summable sequences; you actually see Mellin transformable sequences. Because it follows mellin transform/laplace transform/fourier transfrom rules as old as time.
I mean to say; when I see a Borel summable expansion; I see multiple ways of expanding it in the complex plane; and multiple old techniques from analytic number theory, that just kind of, change things. Totally analytically continue the function.
I believe your two questions were describing and asking of a petal \(\mathcal{P}\)--that it map to a half plane. And largely, if we can define an \(f : P \to P\) with a fixed point on the boundary... I think we're okay.

