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,


\begin{displaymath}
S_{z}
=
\frac{1}{z^{2}-1}\,
C_{z},
\end{displaymath}

where

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


where $k=2j$, 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


\begin{displaymath}
S_{z}
=
\frac{2}{\pi}\,
\frac{z^{2}}{z^{2}-1}
\left[
1
-
\sum_{j=1}^{\infty}
\frac{4}{k(k+2)}\,
z^{k}
\right],
\end{displaymath}

where $k=2j$. 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


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

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

\begin{eqnarray*}
S_{z}
& = &
\frac{1}{\pi}
\left[
1
-
\mbox{\boldmath$\i...
...{j=1}^{\infty}
\frac{4}{k(k+2)}\,
\cos[(k+1)\theta]
\right\},
\end{eqnarray*}


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


\begin{displaymath}
f_{\rm s}(\theta)
=
\frac{1}{\pi\sin(\theta)}
\left\{
-...
...=1}^{\infty}
\frac{4}{k(k+2)}\,
\cos[(k+1)\theta]
\right\},
\end{displaymath}

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


\begin{displaymath}
f_{\rm c}(\theta)
=
\frac{1}{\pi\sin(\theta)}
\left\{
\...
...=1}^{\infty}
\frac{4}{k(k+2)}\,
\sin[(k+1)\theta]
\right\},
\end{displaymath}

where $k=2j$.