I also stumbled upon this very interesting paper:
http://arxiv.org/pdf/hep-th/9206074
It mentions methods that sum power series in the Mittag-Leffler star. One formula it gives, is this: given a principal branch of an analytic function, represented by its power series \( f(z) = \sum_{n=0}^{\infty} a_n z^n \) at z = 0,
\( f(z) = \int_{0}^{\infty} \exp(-\exp(t)) \sum_{n=0}^{\infty} a_n \frac{(tz)^n}{\mu(n)} dt \)
with
\( \mu(n) = \int_{0}^{\infty} \exp(-\exp(t)) t^n dt \).
But it the formulas don't seem to work when tested numerically. Try it with the reciprocal series for \( f(z) = \frac{1}{1 - z} \), which they mention in the paper, i.e. \( f(z) = \sum_{n=0}^{\infty} z^n \). Then try evaluating using these formulas at \( z = -2 \), which outside the convergence radius for the series, but inside the Mittag-Leffler star. It seems to give huge values. Of course (and I highly suspect this is the case), I've missed something here... what might it be?
ADDENDUM: I see now, this does work... it's just that the terms "hump up" for reasonably large z-values before they get smaller and the thing converges... it needed 768 terms to converge to a few places for z = -2, but got ~0.3333 like I'd expect for \( \frac{1}{1 - -2} = \frac{1}{3} \). I think it gets better after that point since once you're over the "hump" the terms get small fairly quick, I suppose 1024 terms would get much more accuracy but calculating \( \mu(n) \) takes a long time.
http://arxiv.org/pdf/hep-th/9206074
It mentions methods that sum power series in the Mittag-Leffler star. One formula it gives, is this: given a principal branch of an analytic function, represented by its power series \( f(z) = \sum_{n=0}^{\infty} a_n z^n \) at z = 0,
\( f(z) = \int_{0}^{\infty} \exp(-\exp(t)) \sum_{n=0}^{\infty} a_n \frac{(tz)^n}{\mu(n)} dt \)
with
\( \mu(n) = \int_{0}^{\infty} \exp(-\exp(t)) t^n dt \).
But it the formulas don't seem to work when tested numerically. Try it with the reciprocal series for \( f(z) = \frac{1}{1 - z} \), which they mention in the paper, i.e. \( f(z) = \sum_{n=0}^{\infty} z^n \). Then try evaluating using these formulas at \( z = -2 \), which outside the convergence radius for the series, but inside the Mittag-Leffler star. It seems to give huge values. Of course (and I highly suspect this is the case), I've missed something here... what might it be?
ADDENDUM: I see now, this does work... it's just that the terms "hump up" for reasonably large z-values before they get smaller and the thing converges... it needed 768 terms to converge to a few places for z = -2, but got ~0.3333 like I'd expect for \( \frac{1}{1 - -2} = \frac{1}{3} \). I think it gets better after that point since once you're over the "hump" the terms get small fairly quick, I suppose 1024 terms would get much more accuracy but calculating \( \mu(n) \) takes a long time.
