Weierstrass sigma functionWeierstrassSigmaFunction2003-08-25 14:37:102003-08-25 14:37:10DefinitionWeierstrass sigma functionWeierstrass zeta functionWeierstrass eta functionWeierstrasssigmaetazeta% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\Rats}{\mathbb{Q}}\begin{defn}
Let $\Lambda\subset\Complex$ be a lattice. Let $\Lambda^{\ast}$
denote $\Lambda-\{ 0 \}$.
\begin{enumerate}
\item The \emph{Weierstrass sigma function} is defined as the
product
$$\sigma(z;\Lambda)=z\prod_{w\in\Lambda^{\ast}}\left(1-\frac{z}{w}\right)e^{z/w+\frac{1}{2}(z/w)^2}$$
\item The \emph{Weierstrass zeta function} is defined by the sum
$$\zeta(z;\Lambda)=\frac{\sigma'(z;\Lambda)}{\sigma(z;\Lambda)}=\frac{1}{z}+\sum_{w\in\Lambda^{\ast}}\left( \frac{1}{z-w}+\frac{1}{w}+\frac{z}{w^2}\right)$$
Note that the Weierstrass zeta function is basically the
derivative of the logarithm of the sigma function. The zeta
function can be rewritten as:
$$\zeta(z;\Lambda)=\frac{1}{z}-\sum_{k=1}^{\infty}\mathcal{G}_{2k+2}(\Lambda)z^{2k+1}$$
where $\mathcal{G}_{2k+2}$ is the Eisenstein series of weight
$2k+2$.
\item The \emph{Weierstrass eta function} is defined to be
$$\eta(w;\Lambda)=\zeta(z+w;\Lambda)-\zeta(z;\Lambda), \text{for
any } z\in\Complex$$ (It can be proved that this is well defined,
i.e. $\zeta(z+w;\Lambda)-\zeta(z;\Lambda)$ only depends on $w$).
The Weierstrass eta function must not be confused with the
Dedekind eta function.
\end{enumerate}
\end{defn}