First-Order Center Series

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

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

where

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

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 $1/k$, this series has coefficients that go to zero as $1/k^{2}$ 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

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

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

$$\frac{z}{z^{2}+1} = \frac{1}{2\cos(\theta)},$$

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

$$S_{z} = \frac{2}{\pi\cos(\theta)} \left\{ 1 + \sum_{j... ...fty} \frac{2(-1)^{j}}{k(k+2)}\, \sin[(k+1)\theta] \right\},$$

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

$$f_{\rm c}(\theta) = \frac{2}{\pi\cos(\theta)} \left\{ 1... ...fty} \frac{2(-1)^{j}}{k(k+2)}\, \cos[(k+1)\theta] \right\},$$

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

$$f_{\rm s}(\theta) = \frac{2}{\pi\cos(\theta)} \left\{ \... ...fty} \frac{2(-1)^{j}}{k(k+2)}\, \sin[(k+1)\theta] \right\},$$

where $k=2j+1$.