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<record version="16" id="6434">
 <title>proof of Jacobi's identity for $\vartheta$ functions</title>
 <name>ProofOfJacobisIdentityForVarthetaFunctions</name>
 <created>2004-10-29 23:46:38</created>
 <modified>2005-02-19 14:20:19</modified>
 <type>Proof</type>
<parent id="6427">Jacobi's identity for $\vartheta$ functions</parent>
 <selfproof>0</selfproof>
 <creator id="6075" name="rspuzio"/>
 <author id="6075" name="rspuzio"/>
 <classification>
	<category scheme="msc" code="33E05"/>
 </classification>
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 <content>We start with the Fourier transform of $f(x) = e^{i \pi \tau x^2 + 2 i x z}$:
 $$\int_{-\infty}^{+\infty} e^{i \pi \tau x^2 + 2 i x z} e^{2 \pi ixy} \, dx = (- i \tau)^{-1/2} e^{-i {(z + \pi y)^2 \over \pi \tau}}$$

Applying the Poisson summation formula, we obtain the following:
 $$\sum_{n=-\infty}^{+\infty} e^{i \pi \tau n^2 + 2 i n z} = (- i \tau)^{-1/2} \sum_{n=-\infty}^{+\infty} e^{-i {(z + \pi n)^2 \over \pi \tau}}$$

The left hand \PMlinkescapetext{side} equals $\vartheta_3 (z \mid \tau)$.  The right hand \PMlinkescapetext{side} can be rewritten as follows:
 $$\sum_{n=-\infty}^{+\infty} e^{-i {(z + \pi n)^2 \over \pi \tau}} = e^{-i {z^2 \over \pi \tau}} \sum_{n=-\infty}^{+\infty} e^{-i {\pi n^2 \over \tau} - {2 i n z \over \tau}} = e^{-i {z^2 \over \pi \tau}} \vartheta_3 (z / \tau  \mid -1 / \tau)$$

Combining the two expressions yields
 $$\vartheta_3 (z \mid \tau) = e^{-i {z^2 \over \pi \tau}} \vartheta_3 (z / \tau  \mid -1 / \tau)$$</content>
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