# Weierstrass zeta & sigma fncts, pseudo-periodicity identity

1. Apr 23, 2017

### binbagsss

1. The problem statement, all variables and given/known data

Let ${w_1,w_2}$ be a basis for $\Omega$ the period lattice.

Use $\zeta (z+ w_{i})=\zeta(z)+ n_i$ , $i=1,2$ ; $m \in N$ for the weierstrass zeta function to show that

$\sigma ( z + mw_i )=(-1)^m \exp^{(mn_i(z+mwi/2))}\sigma(z)$

2. Relevant equations

To note that $\sigma(z)$ is an odd function.

$\zeta(z)=\frac{\sigma'(z)}{\sigma(z)}$

3. The attempt at a solution

I am pretty close but messing up with not getting $(-1)^m$

From $\zeta (z+ w_{i})=\zeta(z)+ n_i$ I get $\zeta (z+ mw_{i})=\zeta(z)+ mn_i$ .
$\frac{d}{dz} \log \sigma(z+mw_i) = \frac{d}{dz} \log \sigma(z) + mn_i$
$\sigma(z+mw_i) = \sigma(z)\exp{mn_i z} A$ , $A$ a constant of integration.

And now to determine $A$ I use the oddness of $\sigma(z)$ (odd as in odd function..) by setting $z=\frac{-mw_i}{2}$:
$\sigma(\frac{mw_i}{2})=\sigma(\frac{-mw_i}{2})A\exp^{-m^2n_iw_i/2}$
$\implies A= - \exp^{\frac{m^2n_iw_i}{2}}$
$\sigma(z+mw_i) = - \sigma(z)\exp{mn_i (z+mw_i/2)}$
So I have a minus sign rather than $(-1)^m$ could someone please tell me what I have done wrong?

2. Apr 28, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. May 20, 2017

### binbagsss

well that is rather courteous of you, thank you- you beautiful bot bruv' .