Half-iterate exp(z)-1: hypothese on growth of coefficients
#28
(08/12/2022, 10:53 PM)JmsNxn Wrote:
(08/12/2022, 05:41 AM)Leo.W Wrote:
(08/11/2022, 08:52 PM)ottfried Wrote:
(08/11/2022, 07:03 PM)Leo.W Wrote: 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.

Nope, James, the cauchy integral works. f(z) here refers not to the divergent sum but a truly holomorphic function around z=0. f(z) is analytic around 0, and has a nearest branch cut at z=-1. Btw the integral isn't the point, the last formula is, and this holds true for arbitrary n as f(z) a divergent series.
\[a_n=[z^{-1}]\big(\frac{(e^z-1)f'(z)}{f(z)^{n+1}}\big)\]
For example, we set initially by our knowledge \(f(z)=z+\frac{z^2}{4}+\frac{z^3}{48}+O(z^4)\), and we compute
\(\frac{(e^z-1)f'(z)}{f(z)^{1+1}}=\frac{1}{z}+\frac{1}{2}+\frac{z}{8}+O(z^2)\) agree with \(a_1=1\)
\(\frac{(e^z-1)f'(z)}{f(z)^{2+1}}=\frac{1}{z^2}+\frac{1}{4z}+\frac{1}{24}+O(z^1)\) agree with \(a_2=\frac{1}{4}\)
There surely are self-references, however it allows one to write a recurrence formula for a_n.
Firstly we assume \(f(z)=\sum_{n\ge1}{a_nz^n}\) with\(a_1=1\), and then we have \(f'(z)=\sum_{n\ge0}{(n+1)a_{n+1}z^n}\), with more computation \((e^z-1)f'(z)=\sum_{n\ge1}{c_nz^n}\),
we have \( c_n=\sum_{k=1}^n{\frac{ka_k}{(n+1-k)!}} \) and after more calculation we'll get some \(a_n=T(a_n,a_{n-1},\cdots,a_1)\)
Due to Faa di bruno's formula this a_n formula should be hard to expand.
We can develop many other formulae by different integral transformation, I didnt find an approps one
Here's a complex plot of f(z), showing its holomorphity. I defined a artificial branch cut at \(Re(z)<0 \wedge Im(z)=\pm1\) and a 2D real-to-real plot


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Regards, Leo Smile
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RE: Parabolic Formal Powerseries - by tommy1729 - 09/09/2022, 12:24 AM

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