Improving convergence of Andrew's slog

[jaydfox]

In order to improve convergence, what I've done is taken the coefficients for the two known singularities at 0.31813... + 1.3372357i and its conjugate, and put them in the B vector, rather than 0's. These coefficients aren't the true values from A, but they're much, much, much closer to the true values. This reduces the errors in the final solution by several orders of magnitude.

Does anybody know what JayD meant here with "coefficients for the two known singularities", and how he arranges them and their conjugates in the B vector?

Add the power series for the two known singularities (i.e., the singularities assumed to be closest to the origin, and hence whose terms would limit the radius of convergence): \( \log_c(z-c) \) and \( \log_{\bar{c}}(z-\bar{c}) \)

If my math isn't wrong, the power series for \( \log_c(z-c) \) is \( a_0 + \sum_{k=1}^{\infty} \frac{-x^k}{c^{k+1}} \)

UPDATE: Yeah, my math was right, but I wrote it down wrong. Here's what I meant (and which I later confirmed):
The power series for \( \log_c(z-c) \) is \( a_0 + \sum_{k=1}^{\infty} \frac{-x^k}{k\,c^{k+1}} \)

The first few terms of which are:

Code:

a_0 +
(0.472565386708 + 0.238338175183*I) x +
(0.124126845264 - 0.147164806491*I) x^2 +
(-0.0555043196549 - 0.075086914615*I) x^3 +
(-0.0468665210317 + 0.0199804139488*I) x^4 + ...

Ignoring the value of \( a_0 \) for now, we can add this series and its conjugate to get the following series. This is the combined taylor series of the two logarithmic singularities at the primary fixed points (i.e., added together):

Code:

a_0 +
0.945130773416 x +
0.248253690529 x^2 +
-0.11100863931 x^3 +
-0.0937330420633 x^4 +
0.0100000104867 x^5 +
0.0358794547132 x^6 +
0.00657595348982 x^7 +
-0.0123046860018 x^8 +
-0.00639023591838 x^9 +
0.00327323081386 x^10 +
0.00376926734556 x^11 +
-0.000280141200757 x^12 + ...

The SAGE code I used to generate these (low-precision) numbers is:

Code:

var('a, x');
f(a) = taylor(log(x-a)/a, x, 0, 13);
A = f(0.3181315052047641353 + 1.337235701430689409*I) + f(0.3181315052047641353 - 1.337235701430689409*I);
print A.coefficients()[1:13];

Note that I have log(x-a)/a, rather than log(x-a)/log(a), since a=log(a) and the former computes (slightly) faster.