Funny pictures of sexps

[jaydfox]

As far as I remember you only suggested that for recentering. Which I dont do, I thought that became clear now. The development slog at 0, sexp at -1, behaves like expected with convergence radius 1. So there is no "of course" if I do the same procedure just with a conjugated function.

Yes, sorry, I sometimes am not clear, or perhaps ambiguous is the correct term. There is another reason you must truncate, regardless of whether you've recentered, which is that only the first so many terms of the calculated slog taylor series will be "accurate". As such, you can only use that many terms when reverting the series. As I discussed when originally analyzing Andrew's slog, only about the first half of the terms will be accurate.

To be more precise, for the unaccelerated islog, only about the first half of the terms will seem to converge on the correct values, while the rest will be quite wrong, though of roughly the correct magnitude. (To see this, calculate a system twice as large, and note that only the first half of the original terms will roughly match the first quarter of the new terms.)

For the accelerated slog, the values will of course converge on the taylor series for logarithmic singularities at the fixed points, but only about the first half of the terms of the "residue" (as I called it) will seem to converge on the correct values. Again, to see this, calculate a system twice as large, and note that the entire original series will seem to converge on the first half of the new series, but when you look at the "residue", only the first half of the original series will roughly equal the first quarter of the new series.

(And in all cases, in the absence of a proof that the infinite system is solvable, "correct" means the converging values that we seem to get as we use larger and larger matrices, so we can only analyze matrices much smaller than the largest we have yet managed to solve.)

For the unaccelerated islog, I haven't analyzed how many terms you can successfully revert, though it's probably much less than half. For the accelerated islog, it's just about half, give or take, and I don't have my original notes on it, so I can't be more specific without redoing my calculations. It may have been more than halfway and I was just being conservative, but I remember for sure that 900 terms out of 1200 just got way off, and that was with the accelerated solution.

Count "numerically" as a singularity? It's not a singularity, so at best it means that you just need more terms in the power series.

Of course that was what I was saying. Little changes in the intput cause big changes in the output that means increasing the precision, doesnt it?

Ah, sorry.