Half-iterate exp(z)-1: hypothese on growth of coefficients

[JmsNxn]

And then \(O(a^{2^n})\) is not Borel summable even after k times, but still summable by contour integrals with Residue theorem. And hence we can sum \(O(^n10)\) like \(1-10+10^{10}-10^{10^{10}}+\cdot\)

Ok, this goes now a bit off topic here, but have you seen my evaluations of that alternating series? That has been ...

\[a_n=\frac{1}{2\pi i}\int_{\gamma}{\frac{f(\tau)\mathrm{d}\tau}{\tau^{n+1}}}\]

This doesn't converge. \(f\) isn't holomorphic in a neighborhood of zero, that's the whole discussion here. How close it is to being analytic. \(f\) is analytic in the left half plane, and is only continuous as \(z \to 0\), where it tends to \(0\). You can't take a jordan curve about \(0\).

This Cauchy integral stuff doesn't really help.

Also, as to the note, we are trying to prove it is \(1\)-borel summable, as you are describing, so the coefficients are going something like \(c^nn!\). That's the goal of the theorem, which as we see through numerical tests, that yes, the numbers seem to be pointing towards this. I'm well aware we can borel sum much larger coefficients, but that's not the goal of this theorem; this is specific to the growth of these coefficients--which passing through Borel summation seems like a clean way to approach it.

if i recall correctly , convergeance on the boundary ( in general ) can be an undecidable problem even if we know boundaries on the taylor coefficients.

That would be related to number theory or analytic number theory usually ... or always ?

forgot the details ...

just a thought.


regards

tommy1729

Yeah this is very related to number theory, which is why I was so confident at first   But then I realized this specific problem proves to be just a bit too hard for the tools I'm used to. But I think I see a proof, but I'm not 100%, I might do a more detailed write up in a bit. Because this has proven to be very interesting. The key appears to be something I never noticed before; that:

\[
\sum_{n=0}^N d_n z^n + \vartheta(z) z^{N+1} = g(z)\\
\]

And we can make this approximation as accurate as we want. Which is: for all \(|z| < \delta\) and \(\Re(z) < 0\) and for all \(\epsilon>0\), there exists \(N\) and \(\vartheta\) such that:

\[
|g(z) - \sum_{n=0}^N d_n z^n + \vartheta z^{N+1}| < \epsilon\\
\]

(Note, these choices of \(N\) and \(\vartheta\) depend on \(z\) and \(\epsilon\)--otherwise we'd have holomorphy).

This is detailed in Knopp's book, and I had never seen that before, so I'm trying to make sense of how we can use that \(g(z)\) is continuous as \(z \to 0\), to pull out a \(n!\), through the binomial coefficients. Which should say that \(d_n = O(c^nn!)\). I see a general shape of the argument but I can't nail it yet.