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

[tommy1729]

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 ...

Sorry my focus has been abberated, I had a quick idea, maybe we can prove the asymp of terms in sum by simply residue theorem:
\[f(z)=\sum_{n\ge1}{a_nz^n}, f^2(z)=e^z-1\]
\[a_n=\frac{1}{2\pi i}\int_{\gamma}{\frac{f(\tau)\mathrm{d}\tau}{\tau^{n+1}}}\]
where \(\gamma\) denotes some 1-time winded curve winding the point 0. (can't recall any specific name though)
then by transform and assuming that f(z) behave quite like z around 0 by asymp, we have
\[a_n=\frac{1}{2\pi i}\int_{\gamma}{\frac{(e^t-1)f'(t)\mathrm{d}\tau}{f(t)^{n+1}}}\]
\[a_n=[t^{-1}]\bigg(\frac{(e^t-1)f'(t)}{f(t)^{n+1}}\bigg)\]
expanding \(e^t-1,f'(t),f(t)^{n+1}\) and allocate the coefficients of \([t^{-1}]\) we can arrange and get a reccurence.
Then the recurrence should prove the asymp of a_n.

ok ... but that still has alot of self-reference it seems to me.

there might be a way around it though ... but i dont see it ...

Also - as i kinda pointed out - assuming it is linear and analytic and yet being neither might be an issue when using theorems that normally require exactly that !!
How do i even know it will not give a close to linear function that is analytic in some radius ?

Do not get me wrong , I love the idea.


regards

tommy1729