While trying to solve a Ramanujan integral, it seems that we need to establish this $q$-series identity: $$1 + \sum_{n = 1}^{\infty}\frac{(-1)^{n}q^{n^{2}}}{(1 - q^{2})(1 - q^{4})\cdots(1 - q^{2n})} = \frac{(1 - q^{2})(1 - q^{4})(1 - q^{6})\cdots}{1 + q + q^{3} + q^{6} + q^{10} + \cdots}$$ Is there any simple proof available for this?
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Well this one was easy. I wonder how I could miss this. From Euler's expansion of $(a;q)_{\infty}$ we have $$(a;q)_{\infty} = \sum_{n = 0}^{\infty}\frac{(-1)^{n}q^{n(n - 1)/2}a^{n}}{(q;q)_{n}}$$ Replacing $q$ by $q^{2}$ and then putting $a = q$ we get $$\sum_{n = 0}^{\infty}\frac{(-1)^{n}q^{n^{2}}}{(q^{2};q^{2})_{n}} = (q;q^{2})_{\infty}$$ and thus we need to establish that $$1 + q + q^{3} + q^{6} + \cdots = \frac{(q^{2};q^{2})_{\infty}}{(q;q^{2})_{\infty}}$$ which is the most basic property of Ramanujan function $$\psi(q) = 1 + q + q^{3} + q^{6} + \cdots$$
Paramanand Singh
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Quick question. Does q-series solve any equations besides possibly polynomials equations? Thanks – Тyma Gaidash Jun 07 '22 at 02:34
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1@TymaGaidash: well most famous $q$ series (ie theta functions) satisfy various algebraic equations. I haven't seen them in context of other equations. – Paramanand Singh Jun 07 '22 at 03:30
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There is a treta function question and there are theta functions; which do you mean? – Тyma Gaidash Jun 08 '22 at 01:33
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1@TymaGaidash: theta functions introduced by Jacobi. I updated my previous comment. Let's blame my keyboard app for the problem. – Paramanand Singh Jun 08 '22 at 01:35