stla

I've found a workaround: ``` library(shiny) library(rhandsontable) library(htmlwidgets) ui

Hello @Avatoo It looks like this package is not actively maintained. Use `as.list` and this will work even after the fix will be done. Cheers.

It is implemented in the R package [elliptic](https://github.com/RobinHankin/elliptic/blob/master/R/elliptic.R#L1208). Does it help?

I seriously believe this formula is wrong. This is much more complicated. This took me a long time but I've finally done an implementation. I compared the printed results with...

And here is the Weierstrass Zeta function: ``` def zeta_weierstrass(g2, g3): r1, r2, r3 = polyroots([4, 0, -g2, -g3]) a = sqrt(r1 - r3) b = sqrt(r1 - r2) c...

Here is a Costa surface. It is made with these two Weierstrass functions.  ```python import pyvista as pv import numpy as np e1 = p_weierstrass_from_w1_w2(1/2, 1j/2)(1/2).real c = 4*e1**2...

And here is the Weierstrass "P prime" function (derivative of the P function). ```python def pprime_weierstrass_from_g2_g3(g2, g3): r1, r2, r3 = polyroots([4, 0, -g2, -g3]) a = sqrt(r1 - r3)...

Cleaner code [here](https://gist.github.com/stla/d771e0a8c351d16d186c79bc838b6c48).

As compared to [Wikipedia](https://en.wikipedia.org/wiki/Weierstrass_elliptic_function#Relation_to_Jacobi's_theta_functions), a factor pi is missing in your formula. I'm going to try.

I understand now!! This equality is correct. Using my implementation, one gets it with $$\omega_1=1/2 \quad \text{and} \quad \omega_2=\tau/2.$$ I missed the one-half factor! Now I will try to understand...