4

Let $q$ be a complex number with $|q| < 1$, prove that $$ \prod_{n \in \Bbb{N} } \left(\dfrac{1-q^n}{1+q^n}\right)^{(-1)^n} = \sum_{n \in \Bbb{Z}} q^{n^2} $$

Not sure if this helps but the LHS can also be written as

$$ \prod_{n\in \Bbb{N}\\\text{n odd}}(1+q^n)(1+q^n+q^{2n}+\cdots)\prod_{n \in \Bbb{N} \\\text{n even}} (1-q^n)(1-q^n+q^{2n}-\cdots) $$

I've tried the method here and the method here, but they both give me nightmares. Is there any good idea?

Jerry Chang
  • 882

1 Answers1

6

Note that $$\begin{align}\prod_{n \in \Bbb{N} } \left(\dfrac{1-q^n}{1+q^n}\right)^{(-1)^n}&= \prod_{n \in \Bbb{N} } \left(\dfrac{1-q^{2n}}{1+q^{2n}}\right)\left(\dfrac{1+q^{2n-1}}{1-q^{2n-1}}\right)\\ &=\prod_{n \in \Bbb{N} } \dfrac{1+q^n}{1+q^{2n}}\dfrac{1-q^{n}}{1-q^{2n-1}}(1+q^{2n-1})\\ &=\prod_{n \in \Bbb{N} } \dfrac{1+q^n}{1+q^{2n}}\prod_{n \in \Bbb{N} }\dfrac{1-q^{n}}{1-q^{2n-1}}\prod_{n \in \Bbb{N} }(1+q^{2n-1})\\ &=\prod_{n \in \Bbb{N} }(1+q^{2n-1})\prod_{n \in \Bbb{N} }(1-q^{2n})\prod_{n \in \Bbb{N} }(1+q^{2n-1})\\ &= \prod_{n \in \Bbb{N} }(1-q^{2n})(1+q^{2n-1})^2\\ &=\sum_{n \in \Bbb{Z}} q^{n^2} \end{align}$$ where at the last step we applied the Jacobi triple product for $x=q$ and $y=1$.

Robert Z
  • 145,942
  • 1
    Thank you very much! – Jerry Chang Apr 14 '20 at 11:04
  • Could you also explain the next-to-last step? – user Apr 14 '20 at 12:05
  • 1
    @user We have that $$\prod_{n \in \Bbb{N} } \dfrac{1+q^n}{1+q^{2n}}=\dfrac{\prod_{n \in \Bbb{N} }(1+q^n)}{\prod_{n \in \Bbb{N} }(1+q^{2n})}=\dfrac{\prod_{n \in \Bbb{N} }(1+q^{2n})\prod_{n \in \Bbb{N} }(1+q^{2n-1})}{\prod_{n \in \Bbb{N} }(1+q^{2n})}=\prod_{n \in \Bbb{N} }(1+q^{2n-1})$$ Similarly we handel $\prod_{n \in \Bbb{N} } \dfrac{1-q^n}{1-q^{2n-1}}$. – Robert Z Apr 14 '20 at 12:39
  • 1
    Thank you for adding this explanation! – user Apr 14 '20 at 12:53