If we call the fixed point x, then here's a look at the coefficients a_k of Andrew's slog, divided by the real part of x^(k+1), multiplied by k (to effect the derivative), and multiplied by abs(x^2).
Here are a couple more detailed shots. Apologies for the axis labels on this first one, but SAGE's plotting engine has a few deficiencies.
With a few exceptions in the first handful of terms, the values seem to be converging on 1.0579. As it turns out, \( {\Large x}^{-1.057939991157 i} \) is equal to \( \Large{x}^{1.057939991157*{\large \left|\frac{x^{\tiny -i}}{x}\right|} \). In other words, if you start at a point very near the fixed point, then 4.44695 real iterations and -1.05794 imaginary iterations will get you to the same point.
And as it turns out, Andrew's solution would appear to be strongly reliant on complex iterations counts, and this isn't the only evidence. More on that to come.
Here are a couple more detailed shots. Apologies for the axis labels on this first one, but SAGE's plotting engine has a few deficiencies.
With a few exceptions in the first handful of terms, the values seem to be converging on 1.0579. As it turns out, \( {\Large x}^{-1.057939991157 i} \) is equal to \( \Large{x}^{1.057939991157*{\large \left|\frac{x^{\tiny -i}}{x}\right|} \). In other words, if you start at a point very near the fixed point, then 4.44695 real iterations and -1.05794 imaginary iterations will get you to the same point.
And as it turns out, Andrew's solution would appear to be strongly reliant on complex iterations counts, and this isn't the only evidence. More on that to come.
~ Jay Daniel Fox