Section 14: Analyzing the Annular Puiseux Power Series

$$ \newcommand{\bint}{\displaystyle{\int\hspace{-10.4pt}\Large\mathit{8}}} \newcommand{\res}{\displaystyle{\text{Res}}} $$

For more information about this section, see: On the Branching Geometry of Algebraic Functions

The reader is asked to review Sections 10 and 13 for background on this section.

In this section we study the power expansions of annular algebraic branches. In Section 13, we derived the integral expressions for computing the Puiseux expansions of an algebraic branch. We briefly summarize that derivation here. Given an $n$-cycle branch, $w_n(z)$, the power expansion is given by: $$ \begin{equation} \begin{aligned} w_n(z)&= \sum_{k=0}^{\infty} a_k (z^{1/n})^k+\sum_{k=1}^{\infty} \frac{b_k}{\left(z^{1/n}\right)^k} \\ &=A(z)+S(z) \end{aligned} \label{equation:equation001} \end{equation} $$ with $$ \begin{equation} \begin{aligned} a_k&=\frac{1}{2n\pi i} \bint \frac{w_n(z)}{\left(z^{1/n}\right)^{k+n}} dz\\ b_k&=\frac{1}{2n\pi i} \bint w_n(z)\left(z^{1/n}\right)^{k-n} dz.\\ \label{equation:equation4} \end{aligned} \end{equation} $$ and $A(z)$ being the analytic terms of the series and $S(z)$, the singular terms. Or in symmetrical form: $$ \begin{equation} \begin{aligned} a_k&=\frac{1}{2n\pi i} \bint \frac{w_n(z)}{\left(z^{1/n}\right)^{k+n}} dz\\ w_n(z)&=\sum_{p=-\infty}^\infty a_p (z^{1/n})^p. \end{aligned} \label{equation:equation100} \end{equation} $$

We will again using the function studied in sections 10 and 13: $$ \begin{equation} \begin{aligned} f(z,w)&=(-z^2+z^3)\\ &+(-4 z+3 z^2)w\\ &+(-z^3-9 z^4)w^2\\ &+(-2+8 z+4 z^2-4 z^3)w^3\\ &+(6-8 z^2+7 z^3+8 z^4)w^4 \end{aligned} \label{equation:equation20} \end{equation} $$ In Section 10, we computed the branch continuation table for this function. In Section 13, we analyze the integral expression of (\ref{equation:equation100}) in the following form: $$ \begin{equation} \begin{aligned} c_k&=\frac{1}{2n\pi i} \bint \frac{w_n(t)}{\left(z^{1/n}\right)^{k+n}} dz=\frac{1}{2 n \pi} \int_{t_0}^{t_e} \frac{w_n(t)re^{it}}{\left(re^{it}\right)^{\frac{k+n}{n}}}\\ &=\frac{1}{2 n\pi} \int_{t_0}^{t_e} w_n(t) \left(re^{it}\right)^{-k/n} dt\\ &=\frac{1}{2 n\pi r^{k/n}} \int_{t_0}^{t_e} w_n(t)\big[\cos(tk/n)-i\sin(tk/n)\big] dt\\ &=\frac{1}{2 n\pi r^{k/n}} I(k,n). \end{aligned} \label{eqn5} \end{equation} $$

And in this section, we study Equation $\ref{equation:equation001}$.

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