Talk:Aufgaben:Problem 7
I think there is a small mistake in Solution a) 8. that is if \( a = \frac{1+\omega}{2} \) there is a double zero at \(a \) and no other zeros in P. You can prove this by looking at the derivative
$$ \partial_z (\wp(z)-\wp(a)) \bigg \vert_{z=\frac{1+\omega}{2}} = \partial_z \wp( \frac{1+\omega}{2}) = 0$$
the last equality we know from exercise 9. So the zero at \( a \) must be of order 2. And for the right half of the equation it is obvious from the form
$$-\frac{(z+a)(z-a)}{z^2a^2}\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}\gamma^2 e^{-\frac{2a}{\gamma}}\frac{(z+a-\gamma)(z-a-\gamma)}{(\gamma-z)^2(\gamma-a)^2}$$
that if \( a = \frac{1+\omega}{2} \Rightarrow -a + 1 + \omega = a \) there is a double zero at \( a \)
Also I think the third Claim is proven in the script: Fourier-Heat page 53. But I’m not sure of we are allowed to use it?
