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I was thinking about the following infinite product:

$$\prod_{n=0}^{\infty} \frac{ae^{-2n}+be^{-n}+c}{c}$$

The right way of generalizing it is to think in terms of $q$-Pochhammer symbols. If $r_{1}$ and $r_{2}$ are the roots of $(a/c)x^{2}+(b/c)x+1$, then (with $|q|<1$:

$$\prod_{n=0}^{\infty} \frac{aq^{2n}+bq^{n}+c}{c} = (-r_{1},q)_{\infty}(-r_{2},q)_{\infty}$$

So this got me thinking: what happens if you replace the polynomial with a power series whose constant term is one, for instance, for the cosine, would it be true that:

$$\prod_{n=0}^{\infty} \cos(q^{n}) = \prod_{n=0}^{\infty} \left[\sum_{m=0}^{\infty} \frac{(-1)^mq^{2nm}}{(2m)!}\right] = \prod_{v=1}^{\infty}(\mp(v\pi-\pi/2,q)_\infty$$

And in general, if $G(z)$ is a power series whose constant term is one and whose roots are $\{r_{1,},r_{2},\ldots\}$, would it then be generally true that $$\prod_{n=0}^{\infty} G(q^{n}) = \prod_{v=1}^{\infty} (-r_{v},q)_{\infty}$$ ?

EDIT: I think I asked too soon, the premise that I asked this question under is mistaken, for $r_{1}$ and $r_{2}$ aren't the roots of the quadratic above.

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If the premise was mistaken, do you want to edit or withdraw (i.e., delete) this question? – ᴡᴏʀᴅs Feb 4 at 3:47

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