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Quite a while ago I asked a question about deducing the Jacobi triple product from the $q$-binomial theorem. In fact, from the $q$-binomial theorem $$ \prod_{i=1}^{m-1}(1+xq^i)=\sum_{j=0}^m\binom{m}{j}_q q^{\binom{j}{2}}x^j, $$ one can infer that $$ \prod_{i=1}^s(1+x^{-1}q^i)\prod_{i=0}^{t-1}(1+xq^i)=\sum_{j=-s}^t\binom{s+t}{s+j}_q q^{\binom{j}{2}}x^j $$ by relebaling $j$ and $x$ to $j'$ and $x'$ and substituting $j=s+j'$ and $m=s+t$ and $x=q^{-s-1}x'$ and then dividing by ${x'}^sq^{-\binom{s}{2}}$. This was pointed out to me by user Colin McQuillan.

By letting $s\to\infty$ and $t\to\infty$, how does one infer the Jacobi triple product $$ \sum_{j\in\mathbb{Z}}(-1)^ja^{\binom{j}{2}}x^j=\prod_{i\geq 0}(1-xa^i)(1-x^{-1}a^{i+1})(1-a^{i+1})? $$ Thanks!

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1 Answer 1

You find such a proof on my homepage, Elementare q-Identitäten, chapter 7

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