Second-Order Center Series

There is a single dominant singularity at $z=-1$, so that we must use factors of $(z+1)^{2}$ in the construction of the second-order center series,

$$S_{z} = \frac{1}{(z+1)^{2}}\, C_{z},$$

where

$$\begin{eqnarray*} C_{z} & = & \frac{2}{\pi}\, (z+1)^{2} \sum_{k=1}^{\infty}... ...1}} {\sqrt{k}\sqrt{k+1}\sqrt{k+2}}\,(-1)^{k} z^{k+1} \right], \end{eqnarray*}$$

where we distributed the factor on the series and manipulated the indices of the resulting sums. Unlike the original series, with coefficients that behave as $1/k^{(1/2)}$, this series has coefficients that go to zero as $1/k^{(5/2)}$ when $k\to\infty$ (although this is not immediately obvious), and therefore our evaluation of the set of dominant singularities of $w(z)$ was in fact correct. We have therefore for $S_{z}$ the representation

$$\begin{eqnarray*} S_{z} & = & \frac{1}{\pi}\, \frac{z}{(z+1)^{2}} \left[ \... ... {\sqrt{k}\sqrt{k+1}\sqrt{k+2}}\, (-1)^{k}\, z^{k+1} \right]. \end{eqnarray*}$$

In order to take the real and imaginary parts of $S_{z}$ on the unit circle, we observe now that since $z=\rho\exp(\mbox{\boldmath$\imath$}\theta)$ we have on the unit circle

$$\frac{z}{(z+1)^{2}} = \frac{1}{2[1+\cos(\theta)]}.$$

If we write this in terms of $\theta/2$ we get

$$\frac{z}{(z+1)^{2}} = \frac{1}{4\cos^{2}(\theta/2)}.$$

We also have that

$$\begin{eqnarray*} -2-\left(4-\sqrt{2}\right)z & = & \left[ -2-\left(4-\sqrt{... ... -\left(8-2\sqrt{2}\right)\sin(\theta/2)\cos(\theta/2) \right], \end{eqnarray*}$$

and therefore we have for $S_{z}$ on the unit circle

$$\begin{eqnarray*} S_{z} & = & \frac{1}{4\pi\cos^{2}(\theta/2)} \times \\ ... ...sqrt{k+1}\sqrt{k+2}}\, (-1)^{k}\, \sin[(k+1)\theta] \right\}, \end{eqnarray*}$$

where we collected the real and imaginary terms. The original DP function is given by the imaginary part,

$$\begin{eqnarray*} f_{\rm s}(\theta) & = & \frac{1}{4\pi\cos^{2}(\theta/2)} \... ...}\sqrt{k+1}\sqrt{k+2}}\, (-1)^{k} \sin[(k+1)\theta] \right\}, \end{eqnarray*}$$

and the corresponding FC function is given by the real part,

$$\begin{eqnarray*} f_{\rm c}(\theta) & = & \frac{1}{4\pi\cos^{2}(\theta/2)} \... ...}\sqrt{k+1}\sqrt{k+2}}\, (-1)^{k} \cos[(k+1)\theta] \right\}. \end{eqnarray*}$$