First-Order Center Series

There is a single dominant singularity at $z=-1$, so that we must use a single factor of $(z+1)$ in the construction of the first-order center series,

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

where

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

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 $1/k$, 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

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

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

$$\frac{z}{z+1} = \frac{1}{2} + \frac{\mbox{\boldmath$\imath$}}{2}\, \frac{\sin(\theta)}{1+\cos(\theta)}.$$

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

$$\frac{z}{z+1} = \frac{1}{2} + \frac{\mbox{\boldmath$\imath$}}{2}\, \frac{\sin(\theta/2)}{\cos(\theta/2)},$$

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

$$\begin{eqnarray*} S_{z} & = & \frac{1}{\pi} \left[ 1 + \mbox{\boldmath$\i... ...ty} \frac{(-1)^{k+1}}{k(k+1)}\, \sin[(k+1/2)\theta] \right\}, \end{eqnarray*}$$

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

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

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

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