Theta Functions and Partitions 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.
 A: 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.
