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

Title:
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
Source:
Advances in Mathematics. 191:408-422
Publisher Information:
Elsevier BV, 2005.
Publication Year:
2005
Document Type:
Academic journal Article
File Description:
application/xml
Language:
English
ISSN:
0001-8708
DOI:
10.1016/j.aim.2004.03.013
Rights:
Elsevier Non-Commercial
Accession Number:
edsair.doi.dedup.....7c73b49e153f24f56c292556f00f06df
Database:
OpenAIRE

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.