Second-Order Center Series

There are two dominant singularities, located at $z=1$ and at $z=-1$, so that we must use factors of $(z-1)^{2}(z+1)^{2}=\left(z^{2}-1\right)^{2}$ in the construction of the second-order center series,

$$S_{z} = \frac{1}{\left(z^{2}-1\right)^{2}}\, C_{z},$$

where

$$\begin{eqnarray*} C_{z} & = & \frac{4}{\pi}\, \left(z^{2}-1\right)^{2} \sum... ...\sum_{j=0}^{\infty} \frac{24}{k(k+2)(k+4)}\, z^{k+3} \right], \end{eqnarray*}$$

where $k=2j+1$, and 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 $1/k$, this series has coefficients that go to zero as $1/k^{3}$ when $k\to\infty$, 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

$$S_{z} = \frac{4}{3\pi}\, \frac{z}{\left(z^{2}-1\right)^{... ...um_{j=0}^{\infty} \frac{24}{k(k+2)(k+4)}\, z^{k+3} \right],$$

where $k=2j+1$. 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

$$\frac{z}{\left(z^{2}-1\right)^{2}} = -\, \frac{z^{*}}{4\sin^{2}(\theta)},$$

where $z^{*}z=1$. We also have that

$$\begin{eqnarray*} z^{*}\left(3-5z^{2}\right) & = & 3z^{*}-5z \\ & = & [-2\cos(\theta)] + \mbox{\boldmath$\imath$} [-8\sin(\theta)], \end{eqnarray*}$$

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

$$\begin{eqnarray*} S_{z} & = & \frac{2}{3\pi\sin^{2}(\theta)} \left\{ \cos(\... ...{\infty} \frac{12}{k(k+2)(k+4)}\, \sin[(k+2)\theta] \right\}, \end{eqnarray*}$$

where $k=2j+1$, and where we collected the real and imaginary terms. The original DP function is given by the imaginary part,

$$f_{\rm s}(\theta) = \frac{2}{3\pi\sin^{2}(\theta)} \left... ...infty} \frac{12}{k(k+2)(k+4)}\, \sin[(k+2)\theta] \right\},$$

where $k=2j+1$, and the corresponding FC function is given by the real part,

$$f_{\rm c}(\theta) = \frac{2}{3\pi\sin^{2}(\theta)} \left... ...infty} \frac{12}{k(k+2)(k+4)}\, \cos[(k+2)\theta] \right\},$$

where $k=2j+1$.