Derivative of a periodic function isn't periodic?

The Weierstrass P-function $$\wp=\wp(z;g_2,g_3)$$ is, in the general case, doubly periodic in the first argument, with two periods $2 \omega_1,2 \omega_2$ which are linearly independent over the reals. That is $$\wp(z+2\omega_i;g_2,g_3)=\wp (z;g_2,g_3), \quad i=1,2$$ and $$\Im \left( \omega_2/\omega_1 \right) \neq 0.$$ Differentiating with respect to $g_3$ one finds that $$\frac{\partial \wp}{\partial g_3}(z;g_2,g_3)=\lim_{h \to 0} \frac{\wp(z;g_2,g_3+h)-\wp(z;g_2,g_3)}{h}=\lim_{h \to 0} \frac{\wp(z+2 \omega_i;g_2,g_3+h)-\wp(z+2 \omega_i;g_2,g_3) }{h}=\frac{\partial \wp}{\partial g_3}(z+2 \omega_i;g_2,g_3) ,$$ so that $\partial \wp_/ \partial g_3$ is also periodic, with the same periods as $\wp$. However, numerical experiments suggest otherwise:

In Mathematica, differentiating with respect to $g_3$ results in

$$\frac{\partial \wp}{\partial g_3}(z;g_2,g_3)=\frac{12 g_2 \wp \left(z;g_2,g_3\right){}^2-18 g_3 \wp \left(z;g_2,g_3\right)+\wp '\left(z;g_2,g_3\right) \left(6 g_2 \zeta \left(z;g_2,g_3\right)-9 g_3 z\right)-2 g_2^2}{2 \left(g_2^3-27 g_3^2\right)},$$ which can also be found in Abramowitz and Stegun. To get a specific case, I chose the periods to be $\{2\omega_i,2 \omega_2\}=\left\{1,i\right\}$. From this one can calculate the invariants $g_3,g_3$ as $$g_2=\frac{2}{3} \pi^4 \left( a^8+b^8+c^8 \right) \approx 189.0727201 \\ g_3=\frac{8}{27} \pi^6 \left(a^{12}-\frac{3}{2} a^8 b^4-\frac{3}{2} a^4 b^8+b^{12} \right) \approx 0$$ with $$a=\theta_2(0;\mathrm{e}^{-\pi}) \\ b=\theta_3(0;\mathrm{e}^{-\pi}) \\ c=\theta_4(0;\mathrm{e}^{-\pi}) .$$ One can then can the following distinct numerical values $$\frac{\partial \wp}{\partial g_3}(0.1;g_2,g_3) \approx 3.57235 \times 10^{-6} \\ \frac{\partial \wp}{\partial g_3}(0.1+1;g_2,g_3) \approx -0.52678$$

My question is: Is $\partial \wp/\partial g_3$ periodic in its first argument or not? If it is, how are the periods related to $2\omega_1,2\omega_2$?

Thank you!

• The periods change if you modify the parameters $g_2,g_3$ (since they depend directly on them), so your manipulation is incorrect. – Pedro Tamaroff Aug 3 '17 at 5:42
• Don't forget that the $\omega_i$ are functions of $g_2$ and $g_3$ which makes your displayed equation following "Differentiating with respect to $g_3$" highly problematic. – Lord Shark the Unknown Aug 3 '17 at 5:42
• How do you justify this step?: $\lim_{h \to 0} \frac{\wp(z;g_2,g_3+h)-\wp(z;g_2,g_3)}{h}=\lim_{h \to 0} \frac{\wp(z+2 \omega_i;g_2,g_3)-\wp(z+2 \omega_i;g_2,g_3) }{h}$ – Chris Culter Aug 3 '17 at 5:42
• @ChrisCulter I forgot an $h$ at the right hand side, corrected. Thank you. – user1337 Aug 3 '17 at 5:44
• Would you expect $\frac{d}{da}\sin ax$ to be periodic in $x$? – Professor Vector Aug 3 '17 at 6:10