Intereting Posts

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An intermediate step deduces Jacobi’s triple product identity by taking 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

$$

and deducing

$$

\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

$$

and then letting $s\to\infty$ and $t\to\infty$. I don’t follow the intermediate deduction, what’s the way to see it? (Thank you Colin McQuillan for pointing this out.)

**Much later edit:** 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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Relabel $j$ and $x$ in the second formula to $j'$ and $x'$ respectively. Substitute $j=s+j'$ and $m=s+t$ and $x=q^{-s-1}x'$ into the first formula. Then divide both sides by $x'^s q^{-\binom{s}{2}}$.

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