First-Order Center Series

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

$$S_{z} = \frac{1}{z^{2}-1}\, C_{z},$$

where

$$\begin{eqnarray*} C_{z} & = & \frac{32}{\pi^{3}} \left(z^{2}-1\right) \sum_... ...^{\infty} \frac{6k(k+2)+8}{k^{3}(k+2)^{3}}\, z^{k+1} \right], \end{eqnarray*}$$

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 $1/k^{3}$, this series has coefficients that go to zero as $1/k^{4}$ 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

$$S_{z} = \frac{32}{\pi^{3}}\, \frac{z}{z^{2}-1} \left[ ... ...\infty} \frac{6k(k+2)+8}{k^{3}(k+2)^{3}}\, z^{k+1} \right],$$

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

$$\frac{z}{z^{2}-1} = -\, \frac{\mbox{\boldmath$\imath$}}{2\sin(\theta)},$$

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

$$\begin{eqnarray*} S_{z} & = & -\, \frac{16}{\pi^{3}}\, \frac{\mbox{\boldmat... ...\frac{6k(k+2)+8}{k^{3}(k+2)^{3}}\, \cos[(k+1)\theta] \right\}, \end{eqnarray*}$$

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

$$f_{\rm s}(\theta) = \frac{16}{\pi^{3}\sin(\theta)} \left... ...rac{6k(k+2)+8}{k^{3}(k+2)^{3}}\, \cos[(k+1)\theta] \right\},$$

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

$$f_{\rm c}(\theta) = \frac{16}{\pi^{3}\sin(\theta)} \left... ...rac{6k(k+2)+8}{k^{3}(k+2)^{3}}\, \sin[(k+1)\theta] \right\},$$

where $k=2j+1$.