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We have the definition $$\vartheta(\tau, z) = \sum_{n=-\infty}^\infty e^{\pi i \cdot(n^2 \tau + 2 n z)}$$ and I want to show $\vartheta(\tau, \tfrac{\tau + 1}{2}) = 0$.

Substituting it in I get $$\begin{eqnarray} \vartheta(\tau, \tfrac{\tau+1}{2}) &=& \sum_{n=-\infty}^\infty e^{\pi i \cdot(n^2 \tau + n \tau + n)} \\ &=& \sum_{n=-\infty}^\infty e^{\pi i \cdot(n^2 \tau + n \tau)} \\ &=& \sum_{n=-\infty}^\infty e^{\pi i \cdot(n^2 + n) \cdot \tau} \\ \end{eqnarray}$$

and a hint was given that the summand is invariant under the change $n \mapsto -(n+1)$, indeed $$(n+1)^2 - n - 1 = n^2 + 2n + 1 - n - 1 = n^2 + n$$ so from this I understand that term $n=0$ equals term $n=-1$, term $n=1$ equals term $n=2$ and we can change the sum to:

$$\vartheta(\tau, \tfrac{\tau+1}{2}) = 2 \sum_{n=0}^\infty e^{\pi i \cdot(n^2 + n) \cdot \tau}$$

but I don't see how this is zero.

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1 Answer 1

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I get $$ \vartheta(\tau, \tfrac{\tau+1}{2}) = \sum_{n=-\infty}^\infty e^{\pi i(n^2 \tau + n \tau + n)} = \sum_{n=-\infty}^\infty(-1)^n e^{\pi i(n^2 +n)\tau} $$ and in this sum the $n$ and $-1-n$ terms cancel each other out.

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  • $\begingroup$ Thank you so much! because it's $\pi i$ not $2 \pi i$! $\endgroup$
    – user581023
    Commented Jul 27, 2020 at 10:57

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