How can complex numbers be used to solve infinite product and sum equations?

[Euge]

Here is this week's POTW:

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Let $q$ be a complex number with $\lvert q \rvert < 1$. Show that

$$\prod_{n = 1}^\infty (1 - q^n) \sum_{n = -\infty}^\infty q^{n+2n^2} = \prod_{n = 1}^\infty (1 - q^{2n})^2$$

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Note: Do not worry about arguments of convergence.Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

No one answered this week's problem. You can read my solution below.

Spoiler

Consider Jacobi's triple product identity

$$\sum_{n = -\infty}^\infty z^n q^{n^2} = \prod_{n = 1}^\infty (1 - q^{2n})(1 + zq^{2n-1})(1 + z^{-1}q^{2n-1})$$

Letting $q\mapsto q^2$ and setting $z = q$ in the identity yields

$$\sum_{n = -\infty}^\infty q^{n + 2n^2} = \prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n-1})(1 + q^{4n-3})\tag{1}$$

Since

$$\prod_{n = 1}^\infty (1 - q^{4n-1})(1 + q^{4n-3}) = \prod_{n\; \text{even}} (1 - q^{2n-1}) \prod_{n\; \text{odd}} (1 - q^{2n-1}) = \prod_{n = 1}^\infty (1 - q^{4n-1})$$

then the product in $(1)$ becomes

$$\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{2n-1})\tag{2}$$

By factorization $1 - q^{4n} = (1 - q^{2n})(1 + q^{4n})$, the product in $(3)$ can be expressed

$$\prod_{n = 1}^\infty (1 - q^{2n})(1 + q^{2n})(1 + q^{2n-1})\tag{3}$$

The product in $(3)$ contains all products of even and odd parts of $1 + q^n$, so $(3)$ becomes

$$\prod_{n = 1}^\infty (1 - q^{2n})(1 + q^n)\tag{4}$$

Since $1 + q^n = \frac{1 - q^{2n}}{1 - q^n}$, $(4)$ can be written

$$\prod_{n = 1}^\infty \frac{(1 - q^{2n})^2}{1 - q^n}$$

Thus

$$\sum_{n = -\infty}^\infty q^{n + 2n^2} = \prod_{n = 1}^\infty \frac{(1 - q^{2n})^2}{1 - q^n}\tag{5}$$

Finally, multiplying both sides of $(5)$ by $\prod_{n = 1}^\infty (1 - q^n)$, the result is established.