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We know Jacobi's identity:

$$ \prod_{n=1}^{\infty} (1+q^{2n-1}z)(1+q^{2n-1}z^{-1})(1-q^{2n}) = \sum_{n=-\infty}^{\infty}q^{n^2}z^n. \tag{$|q|<1$}$$

How do you get identity

\begin{align*} \prod_{n=1}^m (1+x^{2n-1}z) &= 1+\sum_{n=1}^m\frac{(1-x^{2m})(1-x^{2m-2})\dotsm(1-x^{2m-2n+2})}{(1-x^{2})(1-x^{4})\dotsm(1-x^{2n})}x^{n^2}z^n \end{align*}

for some deformation of Jacobi's identity?

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1 Answer 1

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Hint: (May not constitute a full answer, perhaps a partial one)

Rewrite your first line as \begin{align} \prod_{k=1}^\infty (1+xq^{2k-1}) &= \sum_{k=0}^\infty\prod_{m=1}^\infty\frac{(1-q^{2m+2k})}{1-q^{2m}}q^{k^{2}}x^k \\ &= \prod_{m=1}^\infty \frac{1}{1-q^{2m}}\sum_{k=0}^\infty\prod_{m=1}^\infty(1-q^{2m+2k})q^{k^{2}}x^k \\ &=\prod_{m=1}^\infty \frac{1}{1-q^{2m}}\sum_{k=- \infty}^\infty\prod_{m=1}^\infty(1-q^{2m+2k})q^{k^{2}}x^k \end{align} Where, in the last line we used the fact that for $k < 0$ $$\prod_{m=1}^\infty(1-q^{2m+2k})=0$$

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