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

[JmsNxn]

Okay, so I'm going to go out on a limb, and assume you mean that:

\[
f(z) = \exp(z) -1\\
\]
is expanded about \(z =0\), right?

I do have a long history of working with Divergent sums and Umbral Calculus and sum and series speed ups. So I'm very very happy to look into this. But I'm confused by what the bounding function is? I apologize, but I'm a tad confused.

I'd suggest an abel summation personally. And I'll explain how to do it. (...)

Hmm, I think there must be an error somewhere; for the powerseries for fractional iteration of \(f(z)=\exp(z)-1 \) the convergence-radius is not a small \( \delta \)-environment around zero, but exactly zero - that's not my discovery but that of I.N.Baker (and I could not yet understand fully his proof); so I think an Abel-summation should not work.    

I learned the term "limiting a divergent series" or "bounding a divergent series" from the handbook of Konrad Knopp; there he described this as a prerequisite to find an appropriate/powerfule-enough summation-procedure at all - and the growthrate of the coefficients of \( f°^h(z) \) must be estimated by his argument.   
That was the problem that I set myself up to solve: find a function \( A(k) \) (where \(k \) indicates the index of the coefficients in \( f°^h(z) \) ) which is always greater than the \( k \)'th coefficient in \( f°^h(z) \).   
After this - decide for the appropriate summation-technique.  If \( A(k) \) has "geometrical" growth (with \(k \) ) then Euler summation is appropriate/powerful enough, if it has "hypergeometric" growth then Borel-, or other powerful summation techniques are needed.           
This is a (sloppy) paraphrase of Knopp's explanations in chap 13 and 14 - and he didn't (if I recall it correctly) show any summation "stronger" than the Borel-summation; so when we possibly have coefficients in \( f°^h(z) \) growing more than factorially with their index, then there is no classical summation in his book to manage this (except perhaps his chapter about the Euler-MacLaurin summation and the use of asymptotic series, the latter which has explicitely been applied in this forum ... ).                

The function \( A(k) \) that I describe here has factorially growth (plus some geometric growth component) so any \( k \)'th coefficient in \( f°^h(z) \) is inside the bounds given by \( A(k) \) . What I've simply documented in my picture are the values \( a_k \) (and their structural expression) . Then -to have a sanity check- the quotient of the k'th coefficient in \( f°^h(z) \) with \( a_k \). The amazing thing in this is, that my proposal for the \(a_k\) seem lead to an upper limit in the same way as a sinus-curve has one.                 

Well - maybe this can be expressed far better, or even simpler. The literature I had was only Euler, Hardy, Knopp  and then some diverse articles, but no coherent or modern course...  so don't mind to correct me if I've got something basically wrong, I'd like it much if I could settle my experimental results into appropriate form someday :-) ...

Gottfried

Oh no I'm well aware it's exactly zero. Otherwise the series wouldn't be a divergent series!!!!


What I'm saying is that if you take a disk \(\mathbb{E}\) centered at \(-\delta\) of radius \(\delta\), then there is a function \(g(z)\) that is a holomorphic functional square root. Abel sum this at \(z = 0 \)!

This should be equivalent to everything you are doing. Then you have more freedom to bound the taylor terms...

So you would have a holomorphic function \(g(z)\) on \(\mathbb{E} = \{z \in \mathbb{C}\,|\,|z+\delta| < \delta\}\), such that:

\[
g(z) = \sum_{k=0}^\infty g_k(\delta) (z+\delta)^n\\
\]

such that \(g(g(z)) = e^z-1\).

Then if you expand this series about \(z=0\) (you can abel sum it), you should get the same divergent series you are talking about. That's what I meant. You should be able to get a \(k!\) bound from that. You'd have to check:

\[
g^N(z) = \sum_{k=0}^N g_k(\delta) (z+\delta)^k = \sum_{k=0}^N d_k z^k\\
\]

But this should definitely happen. Then all you have to do is find out how to bound the \(g_k\)--which translates to a bound on \(d_k\). You can instantly guess that \(g_k = O(1/\delta^k)\), then a \(k!\) will pull out from the binomial expansion.

---

EDIT:

Also, I think you should be able to get a similar integral transform as euler for this. I'd have to double check some literature.

But just as:

\[
e\int_0^1 \frac{e^{-1/x}}{x}\,dx = 1! - 2! + 3! - 4!... = \sum_{k=1}^\infty (-1)^{k+1} k!\\
\]

You should be able to make an integral transform summation. I believe we call this Borel summation now, but I could be mistaken, I forget the names of people a lot, lol. This is largely because Borel (?) summation is a direct continuation of Abel summation--and these expansions always exist. They basically just use the same tricks as euler, but with some added finesse. Give me a bit, I'll get it...

EDIT2:

OKAY! YES! It's borel summation!

OKay, I'm too sleepy now to do it, but I can bound \(d_k = O(k!)\). I'm busy tomorrow, but I believe I can pull that out for you by sunday. I'll double check everything. But this shouldn't be too hard.

Essentially you just take:

\[
\mathcal{B} g(z) = \sum_{k=0}^\infty g_k(\delta)\frac{(z+\delta)^k}{k!}\\
\]

Then you would get some thing close to:

\[
\sum_{k=1}^\infty d_k\frac{z^k}{k!}\\
\]

And you can show that this has a non-trivial radius of convergence, it'll probably be about \(1\), and therefore \(d_k = \mathcal{O}(k!)\). It might be a bit different though. It might be closer to \(\mathcal{O}(b^kk!)\), or something like that. I have to run the numbers.. Too damn lazy and sleepy rn.

But you should be able to get an expansion that looks like:

\[
\int_0^\infty \frac{e^{-t}}{1+tz}\,dt = \sum_{k=0}^\infty (-1)^k k! z^k\\
\]

And that expression defining \(g(z)\) (whatever it may be) should converge for \(\Re(z) < 0\). Jesus it's been a while since I've done a lot of this stuff, so I may be missing some dumb constants and stuff. Give me til sunday.

Regards, I'm going to bed.