TITLE: How to calculate Weierstrass $p$-function by hand?
QUESTION [6 upvotes]: Let $$\wp(z) = \frac{1}{z^2} + \sum_{\omega \in \Lambda \setminus 0} \frac{1}{(z - \omega)^2} - \frac{1}{\omega^2}$$ where $\Lambda$ is the lattice spanned by $1, \mu = e^{\frac{2\pi i}{3}}.$ We know that $\wp'(1/2) = 0$ as $1/2$ is an order 2 point. I want to calculate the zeros of the elliptic curve equation by computing $\wp(1/2)$ but I'm not sure how to do this by hand. It seems doable as the zeros are relatively non-complicated: $1, \mu_3, \mu_3^2.$ It doesn't seem like we can cancel either as the real part of the first summand $\frac{1}{(z - \omega)^2}$ is fractional.

REPLY [4 votes]: This is just an addendum to Chappers' great answer. We may use the Poisson summation formula to express $\wp\left(\frac{1}{2}\right)$ in a couple of (apparently) different ways. Please allow me to use $\omega$ for denoting the third root of unity $e^{2\pi i/3}=\frac{1+i\sqrt{3}}{2}$. We have
$$\begin{eqnarray*}\wp\left(\tfrac{1}{2}\right)&=&4+\sum_{(a,b)\neq(0,0)}\left[\frac{1}{\left(\frac{1}{2}-a-b\omega\right)^2}-\frac{1}{(a+b\omega)^2}\right]\\&=&\frac{2\pi^2}{3}+\sum_{b\geq 1}\sum_{a\in\mathbb{Z}}\left[\frac{1}{\left(\frac{1}{2}-a-b\omega\right)^2}-\frac{1}{(a+b\omega)^2}\right]\\&=&\frac{2\pi^2}{3}+\pi^2\sum_{b\geq 1}\left[\frac{1}{\cos^2\left(\frac{\pi}{2}b+\frac{\pi}{2}b i\sqrt{3}\right)}-\frac{1}{\sin^2\left(\frac{\pi}{2}b+\frac{\pi}{2}b i\sqrt{3}\right)}\right]\end{eqnarray*} $$
and by separating the contributions of even/odd values of $b$ we get
$$\begin{eqnarray*}\frac{1}{\pi^2}\,\wp\left(\tfrac12\right)&=&\frac{2}{3}+\sum_{n\geq 1}\frac{1}{\cosh^2(\pi n\sqrt{3})}+\sum_{n\geq 1}\frac{1}{\sinh^2(\pi n\sqrt{3})}-\sum_{m\geq 0}\frac{1}{\sinh^2(\frac{2m+1}{2}\pi\sqrt{3})}-\sum_{m\geq 0}\frac{1}{\cosh^2(\frac{2m+1}{2}\pi\sqrt{3})}.\end{eqnarray*}$$
These series are related to special values for the complete elliptic integral of the first kind: see, for instance, this recent survey by Ce Xu. If your purpose is just a numerical evaluation of the LHS, you may just consider suitable partial sums in the RHS: the shown series are already pretty fast-convergent and they can be accelerated, too. For instance, by letting $q=e^{-2\pi\sqrt{3}}$ we have
$$ \sum_{n\geq 1}\frac{1}{4\sinh^2(\pi n\sqrt{3}) }=\sum_{n\geq 1}\frac{q^n}{(1-q^n)^2}=\sum_{n\geq 1}\sum_{k\geq 1}k q^{nk}=\sum_{N\geq 1}\sigma(N) q^N, $$
which essentially is the Eisenstein series $G_2(\tau)$ for $\tau=i\sqrt{3}$.