Cheta with base-change: preliminary results

[jaydfox]

The variable fma in that code is the matrix I'm referring to. Hopefully you can tell from the context what Andrew's talking about. Essentially, it gives us a matrix, which we can multiply on the right by a vector of powers of n, to get a vector representing a power series in x. For fractional n, the terms diverge.

Or, we can multiply on the right by a vector of powers of x (i.e., picking a fixed x), and get a power series in n, which is the iteration function. I use this latter approach.

Oh you mean this matrix is only for the computation of the iteration of \( \exp_\eta \)? And this takes this exuberant time?

Well, calculating the matrix once takes a lot of time. Thereafter, using the matrix is quite fast. The problem is, the coefficients can only reasonably be calculated within a small radius about 0. The more terms you want, the smaller the radius. For 199 terms, using the 200x200 matrix, I already need almost 1000 iterations of log(x+1). Otherwise, the last few terms are inaccurate. For 500 iterations, I only use the first 150 terms, as they get wildly inaccurate at about 180 or 190. If I had a 300x300 matrix, I could probably get the first 250 terms with the same 500 iterations. It's hard to be sure without a little more analysis, because the low order terms start to accumulate errors due to the divergence of the series. I suppose that doesn't make much sense without context, so perhaps I'll dedicate a new thread just to that topic.

I initially thought that the inversion of the interpolation matrix took so much time (and not the fma) thatswhy I provided the direct formula via stirling numbers.

Actually, they both take a lot of time. But I only have to calculate the FMA once, and then I just save it to disk and reuse it. If push came to shove, I could spend the week calculating the 300x300 FMA, but it'd be easier if Andrew could do it in Mathematica in a few minutes or hours and convert it to a format I could use in SAGE. Pretty please!

Now with Gottfried's method of deriving the inverse of the stencil matrix, the time spent there can also be saved! I'll experiment with this and see how large of a grid I can manage within my memory budget.

Then it's just a matter of creating the grid points, which itself is a rather time-consuming process... (hundreds of iterations of log(x+1) and exp(x)-1, using tens of thousands of bits of precision...)