Improving convergence of Andrew's slog

[jaydfox]

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 + ...

Code for confirming this:

Code:

d = 0.3181315052047641353 + 1.337235701430689409*I;
C = [[-1/(k*d^(k+1)), k] for k in xrange(1, 13)];
print C;

Adding the two series then yields the following code:

Code:

d = 0.3181315052047641353 + 1.337235701430689409*I;
C = [[-1/(k*d^(k+1)) + -1/(k*d.conjugate()^(k+1)), k] for k in xrange(1, 13)];
print C;

A simple way to increase efficiency, noting that the imaginary parts cancel:

Code:

d = 0.3181315052047641353 + 1.337235701430689409*I;
C = [[(-2/(k*d^(k+1))).real(), k] for k in xrange(1, 13)];
print C;