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Cauchy's identity states that $$ \prod_{i\geq 0}\frac{1-axq^i}{1-xq^i}=\sum_{n\geq 0}\frac{(1-a)(1-aq)\cdots(1-aq^{n-1})}{(1-q)(1-q^2)\cdots(1-q^n)}x^n. $$

Is it possible to somehow derive this identity as a special case of the $q$-binomial theorem? Mathworld references that it follows as a special case, and I thought maybe setting $a=q^k$ for some power $k$ might lead to it, but I can't say for sure.

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It looks suspiciously like one of the things Ramanujan was playing with... –  J. M. Dec 15 '11 at 6:35
    
The q-binomial theorem is the identity you wrote down above. So what do you mean by Cauchy's identity? –  Noud Dec 15 '11 at 12:56
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Okay, I think that I know what you mean. I tried to calculate it, but there is some small error in it. But I think it is worth to share. There is missing some sign, so your homework is to find the missing sign. Here is the solution:

$ \begin{align*} \prod^{n}_{k=1} (1 + yq^k) &= \frac{(-yq;q)_{\infty}}{(-yq^{n+1};q)_{\infty}} \\ &= {}_1 \phi_0(q^{-n}; q, -yq^{n+1}) \\ &= \sum^{\infty}_{m=0} \frac{(q^{-n};q)_m}{(q;q)_m} (-yq^{n+1})^m \\ &= \sum^{n}_{m=0} y^m (-1)^m q^{m(n+1)} \frac{(1-q^{-n})...(1-q^{-n+m-1})}{(1-q)...(1-q^m)} \\ &= \sum^{n}_{m=0} y^m (-1)^m \frac{(q^{n+1}-q)...(q^{n+1}-q^m)}{(1-q)...(1-q^m)} \\ &= \sum^{n}_{m=0} y^m q \cdot q^2 \cdot ... \cdot q^m (-1)^{2m} \prod^{m-1}_{k=0} \frac{(1-q^{n-k})}{(1-q^{k+1})} \\ &= \sum^{n}_{m=0} y^m q^{m(m+1)/2} \frac{(q)_n}{(q)_m(q)_{n-m}} \end{align*} $

TADA! :D

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Thanks Noud. Where exactly does the identity follow from your work above? –  Sooyoung Dec 15 '11 at 16:41
    
Thank you, the identity (the q-binomial theorem) here is $\frac{(-yq;q)_{\infty}}{(-yq^{n+1};q)_{\infty}} = {}_1 \phi_0(q^{-n}; q, -yq^{n+1})$. –  Noud Dec 16 '11 at 10:00
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