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I am asked to verify the identity

$$\prod_{k = 1}^\infty\left[\frac{(1 + x^{2k - 1})(1 - x^{2k})}{(1 - x^{2k - 1})(1 + x^{2k})}\right] = \sum_{k \in \mathbb{Z}}x^{k^2}.$$

I have tried to simplify the expression inside the product on the left and then use Jacobi's triple product identity to show this but I cannot for the life of me simply the expression.

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2 Answers 2

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The LHS is $$\prod_{k=1}^\infty\frac{(1-x^{4k-2})(1-x^{2k})^2}{(1-x^{2k-1})^2(1-x^{4k})} =\prod_{k=1}^\infty\frac{(1-x^{4k-2})^2(1-x^{2k})}{(1-x^{2k-1})^2} =\prod_{k=1}^\infty(1+x^{2k-1})^2(1-x^{2k}) =\sum_{m=-\infty}^\infty z^{m^2}$$ by the Jacobi triple product.

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  • $\begingroup$ I am not sure how you arrived at your first expression $\endgroup$ Commented Dec 5, 2019 at 3:30
  • $\begingroup$ @EthanDeakins Difference of two squares: $1-a^2=(1+a)(1-a)$. $\endgroup$ Commented Dec 5, 2019 at 3:33
  • $\begingroup$ So wouldn't $(1 - x^{2k}) = (1 - x^k)(1 + x^k)$ $\endgroup$ Commented Dec 5, 2019 at 3:41
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    $\begingroup$ $$(1-x^{4k-2})=(1-x^{2k-1})(1+x^{2k-1}),$$ $$(1-x^{4k})=(1-x^{2k})(1+x^{2k}),$$ $$\prod(1-x^{2k})=\prod(1-x^{4k-2})(1-x^{4k})$$ $\endgroup$
    – mr_e_man
    Commented Dec 5, 2019 at 3:42
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Another way is to multiply and divide by $\prod(1+x^{2k-1})$ to get $$\prod_{k\geq 1}\frac{(1+x^{2k-1})^2(1-x^{2k})}{(1-x^{2k-1})(1+x^k)}$$ The denominator can be written as $$\prod_{k\geq 1}(1-x^{2k-1})\cdot\frac{1-x^{2k}}{1-x^k}=\prod_{k\geq 1}\frac{1-x^k}{1-x^k}=1$$ and hence the desired product is equal to $$\prod_{k\geq 1}(1+x^{2k-1})^2(1-x^{2k})=\sum_{k\in\mathbb {Z}} x^{k^2}$$ via Jacobi Triple Product. Often these product manipulations look trivial when one writes out a few terms and does the cancellation / combination of terms.

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