Computing Abel function at a given center andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 11/30/2007, 06:44 PM jaydfox Wrote:Getting back to the Abel matrix for $e^x$, the simple solution to the divergence problem is either to determine a way to calculate with coefficients of the infinite matrix without resorting to approximations, or to start with a system far larger than the system I intend to solve, and move the center in small steps, truncating along the way. Either way, steps smaller than the radius of convergence must be used. Agreed. In fact I've done some thinking along these lines for the analytic continuation of the natural super-logarithm (both the natural Abel function meaning and the natural base-e meaning). We essentially have 2 methods of computing the natural super-logarithm:By series -- constructing series approximations of the natural Abel function of $e^x$ about a given point. By piece -- using a piecewise-defined function in the complex plane that puts all points within the radius of convergence of the super-logarithm about 0. It would be interesting to see how these two methods compare, assuming its not as bad as you say. If what you say is true, then it is impossible to recenter the series to $z=-2+\pi i$ for example, which has a huge radius of convergence, and a series I would really like to know more about. But I'm still not sure, you definitely got me thinking, since I have had difficulty constructing a series about that point, I thought it was just because I didn't have enough accuracy, maybe I was wrong. However, I do have one series that seems to include points outside of the original radius of convergence, which is why I think you might not be right about the recentering. When I find the natural Abel function of $e^{z-1}$ which is the natural Abel function of $e^z$ about z = -1, with the initial condition $\text{slog}_e(-1) \approx -1.6$ then I get the series: $\text{slog}_e(z) \approx -1.6 + \sum_{k=1}^{n}s_k(z + 1)^k$ which is very well-behaved. If this series was not well-behaved, then I would be inclined to agree with you. But since it is well-behaved, I think its possible to recenter to outside the radius of convergence (but you need to take at least 2 steps, since each step must be within the radius of convergence). Andrew Robbins « Next Oldest | Next Newest »

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