Bessel functions and the iteration of \(e^z -1 \)
#8
I just wanted to note I found the paper I wanted to find. This explains much of the construction \(F\) and its bounds.

It's a paper by D. Zagier; (when I was searching I was writing Zaszlov (don't know why my brain made that mistake)). This is about the various ways of expanding Mellin transform using asymptotic expansions. It does so cross \(z = \infty\) and \(z=0\)--where both are relatable by the mapping \(z \mapsto 1/z\).

https://people.mpim-bonn.mpg.de/zagier/f...lltext.pdf

It has a lot of flavour of number theory (which is where Mellin transforms appear more naturally)--but this paper acts as a more casual introduction to the theorems I speedrunned.

Honestly; all of my results can be traced to this paper. Zagier does it way better though!


Rereading this paper...

Honestly this paper answers everything. It even goes on to show that the worst these objects can grow are \(O(m!)\)--which means they are borel Summable. This is a beautiful paper and absolutely vital to the discussion at hand. I think it answers everyone of Gottfried's questions.

..............

Oh yeah baby!

Quote:......

In summary, if φ(t) is a function of t with asymptotic expansions as a sum of powers of t (or of
powers of t multiplied by integral powers of log t) at both zero and infinity, then we can define in a
canonical way a Mellin transform  ̃φ(s) which is meromorphic in the entire s-plane and whose poles
reflect directly the coefficients in the asymptotic expansions of φ(t). This definition is consistent
with and has the same properties (2) as the original definition (1). We end this section by giving
two simple examples, while Sections 2 and 3 will give further applications of the method.

......


This is basically saying, if we have \(g(\infty)\) has an asymptotic expansion (for us, \(g(\infty) = g^{-1}(-1)\)), and \(g(0)\) has an asymptotic expansion; we can define a meromorphic function on a larger domain than I even claimed. Additionally, in order to belong to the space they are talking about--the asymptotic expansions at worse grow like \(O(m!)\). So, if the Mellin transform converges, the asymptotic series at \(0,\infty\) must be at worst \(O(m!)\)--because otherwise, the mellin transform diverges.
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RE: Bessel functions and the iteration of \(e^z -1 \) - by JmsNxn - 08/25/2022, 05:41 AM

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