Difference between revisions of "Talk:Aufgaben:Problem 7"

Revision as of 12:30, 17 June 2015

Did we prove the orthonormality of characters? meaning are you sure we can use them, it would make proof 13 much shorter...

Carl (talk) 14:30, 17 June 2015 (CEST)

@@ Line 1: / Line 1: @@
-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
+Did we prove the orthonormality of characters? meaning are you sure we can use them, it would make proof 13 much shorter...
-$$ \partial_z (\wp(z)-\wp(a)) \bigg \vert_{z=\frac{1+\omega}{2}} = \partial_z \wp( \frac{1+\omega}{2}) = 0$$
+[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 14:30, 17 June 2015 (CEST)
-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?
-"[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 15:36, 11 January 2015 (CET)"