Let $\vartheta(z)=\sum_{n\in\mathbb{Z}}e^{\pi\mathrm{i}n^2z}$ be a thetafunction. I want to show that $\vartheta(z)+\vartheta(z+1)=2\vartheta(4z)$.
After using the definition I only get to this point:
$\vartheta(z)+\vartheta(z+1)=\sum_{n\in\mathbb{Z}}e^{\pi\mathrm{i}n^2z}+e^{\pi\mathrm{i}n^2z}*e^{\pi\mathrm{i}n^2}=\sum_{n\in\mathbb{Z}}e^{\pi\mathrm{i}n^2z}(1+(-1)^n)$. Can somebody help me with the last step. Thank you
You mean $\vartheta(z)=\sum_{n\in\mathbb{Z}}e^{\pi\mathrm{i} n^2 z}$ and $$\vartheta(z)+\vartheta(z+1)= \sum_{n\in\mathbb{Z}}e^{\pi\mathrm{i}n^2 z}(1+e^{\pi\mathrm{i}n^2 }) =\sum_{n\in\mathbb{Z}}e^{\pi\mathrm{i}n^2 z}(1+(-1)^{n^2})$$ $$ = 2\sum_{n\in\mathbb{Z},n \text{ even}}e^{\pi\mathrm{i}n^2 z} = 2\sum_{n\in\mathbb{Z}}e^{\pi\mathrm{i}(2n)^2 z}=2\sum_{n\in\mathbb{Z}}e^{\pi\mathrm{i}n^2 4 z}=2 \vartheta(4z)$$