Second-Order Center Series

There is a single dominant singularity at $z=-1$, so that we must use a factor of $(z+1)^{2}$ in the construction of the second-order center series,


\begin{displaymath}
S_{z}
=
\frac{1}{(z+1)^{2}}\,
C_{z},
\end{displaymath}

where

\begin{eqnarray*}
C_{z}
& = &
\frac{2}{\pi}\,
(z+1)^{2}
\sum_{k=1}^{\infty}...
...}^{\infty}
\frac{4(-1)^{k+1}}{k(k+1)(k+2)}\,
z^{k+1}
\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 $1/k^{3}$ 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


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

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}{(z+1)^{2}}
=
\frac{1}{2[1+\cos(\theta)]}.
\end{displaymath}

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


\begin{displaymath}
\frac{z}{(z+1)^{2}}
=
\frac{1}{4\cos^{2}(\theta/2)}.
\end{displaymath}

We also have that

\begin{eqnarray*}
2+3z
& = &
[2+3\cos(\theta)]
+
\mbox{\boldmath$\imath$}
...
...]
+
\mbox{\boldmath$\imath$}
[6\sin(\theta/2)\cos(\theta/2)],
\end{eqnarray*}


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

\begin{eqnarray*}
S_{z}
& = &
\frac{1}{4\pi\cos^{2}(\theta/2)}
\left\{
-
1...
... \frac{4(-1)^{k+1}}{k(k+1)(k+2)}\,
\sin[(k+1)\theta]
\right\}.
\end{eqnarray*}


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}{4\pi\cos^{2}(\theta/2)}
\le...
...frac{4(-1)^{k+1}}{k(k+1)(k+2)}\,
\sin[(k+1)\theta]
\right\},
\end{displaymath}

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


\begin{displaymath}
f_{\rm c}(\theta)
=
\frac{1}{4\pi\cos^{2}(\theta/2)}
\le...
...frac{4(-1)^{k+1}}{k(k+1)(k+2)}\,
\cos[(k+1)\theta]
\right\}.
\end{displaymath}