Borel summation

[bo198214]

Thank you Gottfried, for giving me some introduction to divergent summation!

Practically using software we deal with truncated series; but the Borel-sum requires that in the terms of the series as well as the coefficient \( x^k \) or \( x^{rk} \) is present, and we know, that for large \( x \) the exponential series increases to huge values before the influence of the denominator begins to diminuish and suppress that \(x^k\) term. So we cannot really compute with arbitrary large \( x \) (which would be required) but only such large values, that our finitely long exponentialseries has dominated that \(x^k \) or \( x^{rk}\) values. So the implementation of this version of the Borel sum is never really perfect

For instance, to sum the halfiterate of the \( \exp(z)-1 \) I needed order \( r=2 \) but could not use \( x \gt 70 \) say, because of having only 256 or 512 transformed terms. Having 1024 terms, one could insert \( x \) with value of around 80 or 100 and can thus achieve more digits precision.                 

I think what you describe they call "weak Borel summation" and it seems it has the same issue as the "strong" integral version, that you need to check how big your x or t should grow.

P.s.: Practically we need only the proof on concept. After we know that Borel-summation can manage the current growthrate of the \( a_k \) or more precisely that of the \( s_k \) we had of course the possibility not to use \( z=1 \) but the -say- 40th iteration towards 0 by negative height getting \( z_{-40} \lt 1e-10 \), use then for the partial sums \( s_k = \sum_{j=0}^k z_{-40}^k \cdot a_k \) and get nearly exact values for the Borel-sum because \( x \) can now go much farther to \( \infty \) .   

But with this approach we might not need Borel summation at all, we can just take a truncation of the divergent series - say \(h_N(x)\) - and set
\begin{align}
h_{\leftarrow}(x)   &= \lim_{n\to\infty} (f^{\circ -n}(h_N(f^{\circ n}(x))))\\
h_{\rightarrow}(x) &= \lim_{n\to\infty} (f^{\circ n}(h_N(f^{\circ -n}(x))))
\end{align}
where \(f(x)=e^x-1\).
I am not completely sure how big N needs to be, but practically you get high precision with N=20 (I guess N=2 would be the minimum).
It is very similar to the hyperbolic half iterate where you have the formula
\[ \lim_{n\to\infty} (f^{\circ -n}(f'(0)^{\frac{1}{2}}(f^{\circ n}(x)))) \]
where so to say only the first coefficient of the formal powerseries of the half-iterate is taken, but which you could improve by taking more coefficients.