I let my computer run over night to calculate 1295 coefficients of the \(e^x-1\) half iterate.
But even with the new 2k-factorial, one has to be very careful about which t is suitable for calculating.
E.g. with
\[ B(t,u) = e^{-u}\sum_{k=0}^{1295} \frac{c_k}{(2k)!}t^k u^{2k} \]
and t=0.05 one has dangerous values like
\(B(0.05,1500)=-3.50337267212933e600\)
Only for t=0.01 I seem to have small values which I checked up to u=1000000.
Which means we only can directly use Borel summation for values \(|z|\le 0.01\), so I did this for the half-iterate \(h\) of \(f(z)=e^z-1\).
Comparing \(h(h(z))-f(z)\) on the circle with radius \(|z|=0.01\) gave a precision of around 17 digits.
This sounds not bad, however if we compare the the positive half-iterate \(h_+\) and the negative half-iterate \(h_+\) - calculated with traditional means - we get:
\begin{align}
h_-(0.01i)&=-0.000024999999999924043598696143166 + 0.0099999791666927083178232643487\;i\\
h_+(0.01i)&=-0.000024999999999924043599282740330 + 0.0099999791666927083178232889256\;i
\end{align}
i.e. they only differ at the 24th digit ... ! The Borel summed value was just too inprecise to decide whether it is the positive or the negative half-iterate on 0.01i:
\begin{align}
h(0.01i)&=-0.000024999999999924035895201845925 + 0.0099999791666927091010608208421\; i
\end{align}
Very disappointing for a whole night's computation ...
But even with the new 2k-factorial, one has to be very careful about which t is suitable for calculating.
E.g. with
\[ B(t,u) = e^{-u}\sum_{k=0}^{1295} \frac{c_k}{(2k)!}t^k u^{2k} \]
and t=0.05 one has dangerous values like
\(B(0.05,1500)=-3.50337267212933e600\)
Only for t=0.01 I seem to have small values which I checked up to u=1000000.
Which means we only can directly use Borel summation for values \(|z|\le 0.01\), so I did this for the half-iterate \(h\) of \(f(z)=e^z-1\).
Comparing \(h(h(z))-f(z)\) on the circle with radius \(|z|=0.01\) gave a precision of around 17 digits.
This sounds not bad, however if we compare the the positive half-iterate \(h_+\) and the negative half-iterate \(h_+\) - calculated with traditional means - we get:
\begin{align}
h_-(0.01i)&=-0.000024999999999924043598696143166 + 0.0099999791666927083178232643487\;i\\
h_+(0.01i)&=-0.000024999999999924043599282740330 + 0.0099999791666927083178232889256\;i
\end{align}
i.e. they only differ at the 24th digit ... ! The Borel summed value was just too inprecise to decide whether it is the positive or the negative half-iterate on 0.01i:
\begin{align}
h(0.01i)&=-0.000024999999999924035895201845925 + 0.0099999791666927091010608208421\; i
\end{align}
Very disappointing for a whole night's computation ...