- Use the product formula for \(\Theta\) to prove:
- The “triangular number” identity
\[ \prod_ {n = 0} ^ {\infty} (1 + x ^ {n}) (1 - x ^ {2 n + 2}) = \sum_ {n = - \infty} ^ {\infty} x ^ {n (n + 1) / 2}, \]
which holds for \(| x | < 1\)
- The “septagonal number” identity
\[ \prod_ {n = 0} ^ {\infty} (1 - x ^ {5 n + 1}) (1 - x ^ {5 n + 4}) (1 - x ^ {5 n + 5}) = \sum_ {n = - \infty} ^ {\infty} (- 1) ^ {n} x ^ {n (5 n + 3) / 2}, \]
which holds for \(| x | < 1\)
Solution
We use the Jacobi triple product in the form \[ \sum_{n\in\mathbb Z}z^n q^{n(n+1)/2} =\prod_{m=1}^\infty(1-q^m)(1+zq^m)(1+z^{-1}q^{m-1}), \qquad |q|<1. \tag{1} \]
For part (a), set \(z=1\) and \(q=x\). Then \[ \sum_{n\in\mathbb Z}x^{n(n+1)/2} =\prod_{m=1}^\infty(1-x^m)(1+x^m)(1+x^{m-1}). \] Since \((1-x^m)(1+x^m)=1-x^{2m}\), this is \[ \prod_{n=0}^\infty(1+x^n)(1-x^{2n+2}), \] which proves the triangular-number identity.
For part (b), in (1) take \[ q=x^5, \qquad z=-x^{-1}. \] The series side becomes \[ \sum_{n\in\mathbb Z}(-1)^n x^{-n}x^{5n(n+1)/2} =\sum_{n\in\mathbb Z}(-1)^n x^{n(5n+3)/2}. \] The product side is \[ \prod_{m=1}^\infty (1-x^{5m})(1-x^{5m-1})(1-x^{5m-4}). \] Putting \(n=m-1\) in the last two factors rewrites this as \[ \prod_{n=0}^\infty (1-x^{5n+1})(1-x^{5n+4})(1-x^{5n+5}), \] which is the septagonal-number identity.