Result: Ramanujan's 'Lost' notebook IX: the partial theta function as an entire function: Ramanujan's ``lost'' notebook. IX: The partial theta function as an entire function
https://www.sciencedirect.com/science/article/abs/pii/S0001870804001008
https://core.ac.uk/display/82347085
https://pennstate.pure.elsevier.com/en/publications/ramanujans-lost-notebook-ix-the-partial-theta-function-as-an-enti
https://dialnet.unirioja.es/servlet/articulo?codigo=1015744
Further Information
In Ramanujan's ``lost'' notebook, it is claimed that \[ \sum_{n=0}^{\infty} a^nq^{n^2} = \prod_{n=1}^{\infty} \left(1 + aq^{2n-1}(1 + y_1(n) + y_2(n) + \cdots) \right), \] where \[ \begin{aligned} y_1(n) &= \frac{\sum_{j \geq n} (-1)^jq^{j^2+j}}{\sum_{j \geq 0} (-1)^j(2j+1)q^{j^2+j}}\\ \text{and} y_2(n) &= \frac{\left(\sum_{j \geq n} (j+1)(-1)^jq^{j^2+j}\right) \left(\sum_{j \geq n} (-1)^jq^{j^2+j} \right) } {\left(\sum_{j\geq 0} (-1)^j(2j+1)q^{j^2+j} \right)^2}. \end{aligned} \] This is reminiscent of the identity treated Part VIII: The entire Rogers-Ramanujan function [Adv. Math. 191, No. 2, 393--407 (2005; Zbl 1067.11062)]. Indeed, Andrews uses more or less the same method to prove the above assertion. One notable difference is that the key polynomials were orthogonal in that paper but are not orthogonal in the present case.