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  "body": "\n## Statement\n\nLet $a, q, z \\in \\C$ with $\\abs{q} \u003c 1$ and $\\abs{z} \u003c 1$.\nThen\n$$\\sum_{n=0}^{\\infty} \\frac{(a; q)_n}{(q; q)_n} z^n = \\frac{(az; q)_\\infty}{(z; q)_\\infty},$$\nwhere $(\\cdot\\,; q)_n$ and $(\\cdot\\,; q)_\\infty$ are [[q-Pochhammer symbols|Y1X8ZZQ7VM]].\nIn the notation of [[basic hypergeometric series|EB4N9RQR51]], ${}_1\\phi_0(a; -; q, z) = \\frac{(az; q)_\\infty}{(z; q)_\\infty}$.\n\n## Remarks\n\nProof idea: let $f(z) = \\sum_n c_n z^n$ be the left side, with $c_n = \\frac{(a; q)_n}{(q; q)_n}$.\nComparing the coefficients of $z^n$ on both sides, with $c_n (1 - q^n) = c_{n-1} (1 - a q^{n-1})$, gives $(1 - z) f(z) = (1 - az) f(qz)$.\nIterating this functional equation gives $f(z) = \\frac{(az; q)_N}{(z; q)_N} f(q^N z)$, and $f(q^N z) \\to f(0) = 1$ as $N \\to \\infty$.\nSpecial cases: $a = 0$ gives $\\sum_n \\frac{z^n}{(q; q)_n} = \\frac{1}{(z; q)_\\infty}$, and replacing $a$ by $a/b$ and $z$ by $bz$ and letting $b \\to 0$ gives $\\sum_n \\frac{(-1)^n q^{\\binom{n}{2}} a^n z^n}{(q; q)_n} = (az; q)_\\infty$.\nWith $a = q^{-N}$ and $z$ replaced by $-q^N z$, one obtains the finite form $\\prod_{k=0}^{N-1} (1 + q^k z) = \\sum_{k=0}^{N} q^{\\binom{k}{2}} \\binom{N}{k}_q z^k$ with the [[Gaussian binomial coefficients|YS09Z1YNEK]].\nThe hypothesis $\\abs{z} \u003c 1$ is needed for $a \\notin \\set{1, q^{-1}, q^{-2}, \\dots}$, since the series then has radius of convergence $1$.\n\n## Explanations\n\n### L1 human\n\nThe binomial series says $\\sum_n \\binom{-a}{n} (-z)^n = (1 - z)^{-a}$.\nThe $q$-binomial theorem is its $q$-analogue: the power $(1 - z)^{-a}$ becomes a ratio of two infinite products.\nLetting $q \\to 1$ with $a$ replaced by $q^{a}$ recovers the [[binomial theorem|0EHWPWGHB2]] for real exponents.\n\n### L2 human\n\nThis is the simplest summation formula for basic hypergeometric series, and many others follow from it.\nTogether with the case $a = 0$ it gives Euler's two expansions of $1/(z; q)_\\infty$ and $(z; q)_\\infty$, which are generating functions for partitions into parts of a given kind.\nIts bilateral extension is [[Ramanujan's 1psi1 summation|RK6DCJVJCR]].\n",
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