Theta Functions and Partitions

Question

I am reading some papers by Ramanujan on congruence properties of the partition function. At one point he says that he will be using "theta functions" and introduces the following:

It can be shewn that $$ \begin{align} &\dfrac{(1-x^5)(1-x^{10})(1-x^{15})\dots}{(1-x^{1/5})(1-x^{2/5})(1-x^{3/5})\dots} = \dfrac{1}{\xi^{-1}-x^{1/5}-\xi x^{2/5}}\\ &= \dfrac{\xi^{-4}-3x\xi+x^{1/5}(\xi^{-3}+2x\xi^2)+x^{2/5}(2\xi^{-2}-x\xi^3)+x^{3/5}(3\xi^{-1}+x\xi^4)+5x^{4/5}}{\xi^{-5}-11x-x^2\xi^5} \end{align} $$ where $$ \xi = \dfrac{(1-x)(1-x^4)(1-x^6)(1-x^9)\dots}{(1-x^2)(1-x^3)(1-x^7)(1-x^8)\dots} $$ the indices of the powers of x, both in the numerator and denominator of $\xi$, forming two arithmetical progressions with common difference 5. It follows that:

$$ (1-x^5)(1-x^{10})(1-x^{15})\dots\{p(4)+p(9)x+p(14)x^2\dots\} = \dfrac{5}{\xi^{-5}-11x-x^2\xi^5} $$

Written a little cleaner he is saying that:

$$ \left(\prod_{n=1}^\infty(1-x^{5n})\right)\left(\sum_{n=0}^\infty p(5n+4)x^{n}\right) = \dfrac{5}{\xi^{-5}-11x-x^2\xi^5} $$

I don't have any experience with this function or theta functions. I would appreciate some references to read more about these theta functions in general, some understanding of why he uses this identity with powers of 1/5 and how it is derived, and help understanding how this is connected to the partition function for these particular values.

Answer 1

The general Ramanujan theta function is defined by $$ f(a,b) := 1 + (a+b) + ab(a^2+b^2) + (ab)^3(a^3+b^3) + \dots. \tag{1} $$ which factors according to the Jacobi triple product as $$ f(a,b) = (-a;ab)_\infty(-b;ab)_\infty(ab;ab)_\infty. \tag{2} $$ An important special case is the single variable theta function $$ f(-x) := f(-x,-x^2) = (1-x)(1-x^2)(1-x^3)\cdots. \tag{3} $$ For convenience define the variable $\, q := x^{1/5} \,$ so that $\, x = q^5.\,$ Define the functions $$ r := \frac{f(-x,-x^4)}{f(-x^2,-x^3)} = \frac{(1-x)(1-x^4)(1-x^6)(1-x^9)\cdots} {(1-x^2)(1-x^3)(1-x^7)(1-x^8)\cdots}, \tag{4} $$ $$ y := f(-x^5)/f(-q),\;\;\text{ and } \;\; z := (f(-x^5)/f(-x))^6. \tag{5} $$ For technical reasons introduce the variants $$ R := q\,r, \quad Y := q\,y, \quad Z := x\,z. \tag{6} $$ Somehow Ramanujan has proved that $$ R\,Y^{-1} = 1 - R - R^2 \tag{7} $$ (which is a series multisection) and also proved that $$ R^5Z^{-1} = 1 - 11\,R^5 - R^{10}. \tag{8} $$ This implies that $$ R^4\,Y\,Z^{-1} = (1-11\,R^5-R^{10})/(1-R-R^2). \tag{9} $$ Dividing the two polynomials gives the result $$ R^4YZ^{-1}=R^8-R^7+2R^6-3R^5+5R^4+3R^3+2R^2+R+1.\tag{10} $$ Divide both sides by $\,R^4\,$ and pair up the powers of $\,R\,$ and $\,q\,$ to get $$ YZ^{-1} \!=\! 5 \!+\! (R^{-4}\!-\!3R) \!+\! (R^{-3}\!+\!2R^2) \!+\! (2R^{-2}\!-\!R^3) \!+\! (3R^{-1}\!+\!R^4) \tag{11} $$ which is a series multisection. Use equations $(5),(6)$ to rewrite this as $$ f(-x^5)q^{-4}f(-q)^{-1}z^{-1} \!=\! A_0\!+\! A_1 \!+\!A_2 \!+\! A_2 \!+\! A_4 \;\text{ where } \;A_0 \!=\! 5. \tag{12} $$ Now also use series multisection to get $$ f(-q)^{-1} = p_0 + p_1 + p_2 + p_3 + p_4 \tag{13} $$ where $$ p_k := \sum_{n=0}^\infty p(5n+k)\,q^{5n+k} = q^k\sum_{n=0}^\infty p(5n+k)\,x^n. \tag{14} $$ Use the multisection equation $(12)$ to select integer powers of $\,x\,$ to get $$ f(-x^5)\,p_4\, z^{-1} = 5. \tag{15} $$ Use equations $(5),(6),(8)$ to get $$ z^{-1} = r^{-5} - 11\,x - x^2\,r^5. \tag{16} $$ The final result is $$ f(-x^5)\,p_4 = \frac5{r^{-5} -11\,x -x^2\,r^5} \tag{17}$$ where $\,r\,$ is denoted by $\,\xi\,$ in Ramanujan's paper.