Elementary Approach to Modular Equations: Ramanujan's Theory 5

Quintuple Product Identity

We first establish an identity similar to Jacobi's Triple Product which involves five factors and is quite useful in establishing various other identities involving q-series and products. This was first introduced in the mathematical literature by G. N. Watson in order to prove some of Ramanujan's theorems. The quintuple product identity is given by $$\begin{align}&\prod_{n = 1}^{\infty}(1 - q^{n})(1 - q^{n}z)(1 - q^{n - 1}z^{-1})(1 - q^{2n - 1}z^{2})(1 - q^{2n - 1}z^{-2})\notag\\ &\,\,\,\,\,\,\,\,= \sum_{n = -\infty}^{\infty}q^{n(3n + 1)/2}(z^{3n} - z^{-3n - 1})\notag\end{align}$$
In more compact form this is written as: $$\begin{align}&(q;q)_{\infty}(qz;q)_{\infty}(1/z;q)_{\infty}(qz^{2};q^{2})_{\infty}(q/z^{2};q^{2})_{\infty}\notag\\ &\,\,\,\,\,\,\,\,\,= \sum_{n = -\infty}^{\infty}q^{(3n^{2} + n)/2}(z^{3n} - z^{-3n - 1})\tag{1}\end{align}$$ If we replace $q$ by $q^{2}$ and $z$ by $z/q$ we get $$\begin{align}& (q^{2};q^{2})_{\infty}(qz;q^{2})_{\infty}(q/z;q^{2})_{\infty}(z^{2};q^{4})_{\infty}(q^{4}/z^{2}; q^{4})_{\infty}\notag\\ &\,\,\,\,\,\,\,\,= \sum_{n = -\infty}^{\infty}q^{3n^{2} + n}(z^{3n}q^{-3n} - z^{-3n - 1}q^{3n + 1})\tag{2}\end{align}$$ It is the above form which we will prove now. Let the infinite product on the left of above equation be denoted by $F(z, q)$ and let's assume that $$F(z, q) = \sum_{n = -\infty}^{\infty}a_{n}z^{n}$$ where $a_{n} = a_{n}(q)$ are dependent only on $q$.

Then $$\begin{align} F(zq^{2}, q) &= (q^{2}; q^{2})_{\infty}(zq^{3};q^{2})_{\infty}(1/zq;q^{2})_{\infty}(z^{2}q^{4};q^{4})_{\infty}(1/z^{2};q^{4})_{\infty}\notag\\ &= (q^{2};q^{2})_{\infty}\frac{(qz;q^{2})_{\infty}}{1 - qz}(1 - 1/qz)(q/z;q^{2})_{\infty}\frac{(z^{2};q^{4})_{\infty}}{1 - z^{2}}(1 - 1/z^{2})(q^{4}/z^{2};q^{4})_{\infty}\notag\\ &= \frac{F(z, q)}{qz^{3}}\notag\end{align}$$ therefore $F(z, q) = qz^{3}F(zq^{2}, q)$. Using this equation and the assumed series expansion of $F(z, q)$ we see that $a_{n} = q^{2n - 5}a_{n - 3}$ and therefore we have $$a_{3n} = a_{0}q^{3n^{2} - 2n},\,\,\, a_{3n + 1} = a_{1}q^{3n^{2}},\,\,\, a_{3n + 2} = a_{2}q^{3n^{2} + 2n}$$ In a similar manner we can prove that $F(z, q) = -z^{2}F(1/z, q)$ and this leads to the fact that $a_{2} = -a_{0}$ and $a_{1} = 0$. Therefore we finally have $$\begin{align} F(z, q) &= a_{0}(q) \sum_{n = -\infty}^{\infty}(q^{3n^{2} - 2n}z^{3n} - q^{3n^{2} + 2n}z^{3n + 2})\notag\\ &= a_{0}(q)\sum_{n = -\infty}^{\infty}q^{3n^{2} + n}(z^{3n}q^{-3n} - z^{-3n - 1}q^{3n + 1})\notag\end{align}$$ Here in the second sum we have replaced the index $n$ by $-(n + 1)$. It now remains to prove that $a_{0}(q) = 1$. By changing the index of summation from $n$ to $(n - 1)$ in the second sum we get $$F(z, q) = a_{0}(q)\sum_{n = -\infty}^{\infty}q^{3n^{2} - 2n}\left(z^{3n} - \frac{1}{z^{3n - 2}}\right)$$ If we put $z = i$ in the above relation and note that $$\begin{align} i^{3n} - \frac{1}{i^{3n - 2}} &= i^{3n} + \frac{i^{-2}}{i^{3n - 2}}\notag\\ &= i^{3n} + i^{-3n} = e^{3n\pi i / 2} + e^{-3n \pi i / 2}\notag\\ &= 2\cos\left(\frac{3n}{2}\,\pi\right)\notag\end{align}$$ then we get $$\begin{align} F(i, q) &= 2a_{0}(q)\sum_{n = -\infty}^{\infty}q^{3n^{2} - 2n}\cos\left(\frac{3n}{2}\,\pi\right)\notag\\ &= 2a_{0}(q)\sum_{n = -\infty}^{\infty}q^{12n^{2} - 4n}(-1)^{n}\notag\end{align}$$ (here we use the fact that the cosine function above vanishes when $n$ is odd and is $(-1)^{n/2}$ when $n$ is even and we then replace $n$ by $2n$)

Again from the product expansion $$\begin{align}F(i, q) &= (q^{2};q^{2})_{\infty}(iq;q^{2})_{\infty}(-iq;q^{2})_{\infty}(-1;q^{4})_{\infty}(-q^{4}; q^{4})_{\infty}\notag\\ &= 2(q^{2};q^{2})_{\infty}(iq;q^{2})_{\infty}(-iq;q^{2})_{\infty}(-q^{4};q^{4})_{\infty}^{2}\notag\\ &= 2(q^{2};q^{2})_{\infty}(-q^{2};q^{4})_{\infty}(-q^{4};q^{4})_{\infty}(-q^{4};q^{4})_{\infty}\notag\\ &= 2(q^{2};q^{2})_{\infty}(-q^{2};q^{2})_{\infty}(-q^{4};q^{4})_{\infty}\notag\\ &= 2(q^{4};q^{4})_{\infty}(-q^{4};q^{4})_{\infty}\notag\\ &= 2(q^{8};q^{8})_{\infty}\notag\\ &= 2\sum_{n = -\infty}^{\infty}(-1)^{n}q^{12n^{2} - 4n}\notag\end{align}$$ (in the last line we have used the Euler's Pentagonal Theorem). Comparing the two expressions of $F(i, q)$ we see that $a_{0}(q) = 1$. Hence the proof of the quintuple product identity is complete.

Using this quintuple product identity we can derive many results some of which will help in finding modular equations. For example we can write the identity in the form $$(q;q)_{\infty}(qz;q)_{\infty}(q/z;q)_{\infty}(qz^{2};q^{2})_{\infty}(q/z^{2};q^{2})_{\infty} = \sum_{n = -\infty}^{\infty}q^{(3n^{2} + n)/2}\,\,\frac{z^{3n} - z^{-3n - 1}}{1 - 1/z}$$ If we let $z \to 1$ we get $$\phi^{2}(-q)f(-q) = (q;q)_{\infty}^{3}(q;q^{2})_{\infty}^{2} = \sum_{n = -\infty}^{\infty}(6n + 1)q^{(3n^{2} + n)/2}\tag{3}$$ Similarly from the identity $$(q^{2};q^{2})_{\infty}(qz;q^{2})_{\infty}(q/z;q^{2})_{\infty}(z^{2};q^{4})_{\infty}(q^{4}/z^{2};q^{4})_{\infty} = \sum_{n = -\infty}^{\infty}q^{3n^{2} + n}(z^{3n}q^{-3n} - z^{-3n - 1}q^{3n + 1})$$ we arrive at the following (by replacing $n$ with $(n + 1)$ in the first sum): $$\begin{align}&(q^{2};q^{2})_{\infty}(qz;q^{2})_{\infty}(q/z;q^{2})_{\infty}(z^{2};q^{4})_{\infty}(q^{4}/z^{2};q^{4})_{\infty}\notag\\ &\,\,\,\,\,\,\,\,= \sum_{n = -\infty}^{\infty}q^{3n^{2} + 4n + 1}(z^{3n + 3} - z^{-3n - 1})\notag\end{align}$$ and therefore $$\begin{align}&(q^{2};q^{2})_{\infty}(qz;q^{2})_{\infty}(q/z;q^{2})_{\infty}(q^{4}z^{2};q^{4})_{\infty}(q^{4}/z^{2};q^{4})_{\infty}\notag\\ &\,\,\,\,\,\,\,\,= \sum_{n = -\infty}^{\infty}q^{3n^{2} + 4n + 1}\,\frac{z^{3n + 3} - z^{-3n - 1}}{1 - z^{2}}\notag\end{align}$$ Taking limits when $z \to 1$ we get $$(q^{2};q^{2})_{\infty}(q;q^{2})_{\infty}^{2}(q^{4};q^{4})_{\infty}^{2} = \sum_{n = -\infty}^{\infty}(-3n - 2)q^{3n^{2} + 4n + 1}$$ Replacing $n$ by $(-n)$ in the sum on the right we get $$(q^{2};q^{2})_{\infty}(q;q^{2})_{\infty}^{2}(q^{4};q^{4})_{\infty}^{2} = \sum_{n = -\infty}^{\infty}(3n - 2)q^{3n^{2} - 4n + 1}$$ Again replacing $n$ by $(n + 1)$ we get $$(q^{2};q^{2})_{\infty}(q;q^{2})_{\infty}^{2}(q^{4};q^{4})_{\infty}^{2} = \sum_{n = -\infty}^{\infty}(3n + 1)q^{3n^{2} + 2n}\tag{4}$$ In terms of Ramanujan's functions we can rewrite above identity as $$\psi(q^{2})f^{2}(-q) = (q^{2};q^{2})_{\infty}(q;q^{2})_{\infty}^{2}(q^{4};q^{4})_{\infty}^{2} = \sum_{n = -\infty}^{\infty}(3n + 1)q^{3n^{2} + 2n}\tag{5}$$

Relation Between Theta Functions of Argument $q$ and $q^{3}$

Using the above identities we will establish relations between functions of argument $q$ and $q^{3}$. To start with we have from equation $(5)$ $$\begin{align}\psi(q^{2})f^{2}(-q) &= \sum_{n = -\infty}^{\infty}(3n + 1)q^{3n^{2} + 2n}\notag\\ &= \frac{d}{dz}\left(\sum_{n = -\infty}^{\infty}q^{3n^{2} + 2n}z^{3n + 1}\right)\arrowvert_{z = 1}\notag\\ &= \frac{d}{dz}\{zf(q^{5}z^{3}, q/z^{3})\}\arrowvert_{z = 1}\notag\\ &= f(q, q^{5})\,\frac{d}{dz}\{\log\{zf(q^{5}z^{3}, q/z^{3})\}\}\arrowvert_{z = 1}\notag\\ &= f(q, q^{5})\left(1 + \frac{d}{dz}\{\log\{(-q^{5}z^{3};q^{6})_{\infty}(-q/z^{3};q^{6})_{\infty}(q^{6};q^{6})_{\infty}\}\}\arrowvert_{z = 1}\right)\notag\\ &= f(q, q^{5})\left(1 + 3\left(\sum_{n = 0}^{\infty}\frac{q^{6n + 5}}{1 + q^{6n + 5}} - \sum_{n = 0}^{\infty}\frac{q^{6n + 1}}{1 + q^{6n + 1}}\right)\right)\notag\end{align}$$ Replacing $q$ with $(-q)$ we get $$\begin{align} 1 + 3\sum_{n = 0}^{\infty}\left(\frac{q^{6n + 1}}{1 - q^{6n + 1}} - \frac{q^{6n + 5}}{1 - q^{6n + 5}}\right) &= \frac{\psi(q^{2})f^{2}(q)}{f(-q, -q^{5})}\notag\\ &= \frac{\psi(q^{2})f^{2}(q)}{(q;q^{6})_{\infty}(q^{5};q^{6})_{\infty}(q^{6};q^{6})_{\infty}}\notag\\ &= \frac{\psi(q^{2})f^{2}(q)(q^{3};q^{6})_{\infty}}{(q;q^{2})_{\infty}(q^{6};q^{6})_{\infty}}\notag\\ &= \frac{\psi^{2}(q)f^{2}(q)}{\phi(q)\chi(-q)\psi(q^{3})}\notag\\ &= \frac{\psi^{2}(q)\phi(q)\psi(-q)}{\psi(q^{3})\chi(-q)f(q)}\notag\\ &= \frac{\psi^{2}(q)\chi(q)\psi(-q)}{\psi(q^{3})\chi(-q)}\notag\\ &= \frac{\psi^{3}(q)}{\psi(q^{3})}\notag\end{align}$$ Thus we arrive at $$\frac{\psi^{3}(q)}{\psi(q^{3})} = 1 + 3\sum_{n = 0}^{\infty}\left(\frac{q^{6n + 1}}{1 - q^{6n + 1}} - \frac{q^{6n + 5}}{1 - q^{6n + 5}}\right)\tag{6}$$ Similarly we have $$\begin{align} \phi^{2}(-q)f(-q) &= \sum_{n = -\infty}^{\infty}(6n + 1)q^{(3n^{2} + n)/2}\notag\\ &= \frac{d}{dz}\left(\sum_{n = -\infty}^{\infty}q^{(3n^{2} + n)/2}z^{6n + 1}\right)\arrowvert_{z = 1}\notag\\ &= \frac{d}{dz}\{zf(q/z^{6}, q^{2}z^{6})\}\arrowvert_{z = 1}\notag\\ &= f(q, q^{2})\,\frac{d}{dz}\{\log\{zf(q/z^{6}, q^{2}z^{6})\}\}\arrowvert_{z = 1}\notag\\ &= f(q, q^{2})\left(1 + \frac{d}{dz}\{\log\{(-q/z^{6};q^{3})_{\infty}(-q^{2}z^{6};q^{3})_{\infty}(q^{3};q^{3})_{\infty}\}\}\arrowvert_{z = 1}\right)\notag\\ &= f(q, q^{2})\left(1 - 6\sum_{n = 0}^{\infty}\left(\frac{q^{3n + 1}}{1 + q^{3n + 1}} - \frac{q^{3n + 2}}{1 + q^{3n + 2}}\right)\right)\notag\end{align}$$ so that $$1 - 6\sum_{n = 0}^{\infty}\left(\frac{q^{3n + 1}}{1 + q^{3n + 1}} - \frac{q^{3n + 2}}{1 + q^{3n + 2}}\right) = \frac{\phi^{2}(-q)f(-q)}{f(q, q^{2})}$$ Replacing $q$ by $(-q)$ we get $$1 + 6\sum_{n = 0}^{\infty}\left(\frac{q^{6n + 1}}{1 - q^{6n + 1}} - \frac{q^{6n + 4}}{1 + q^{6n + 4}} + \frac{q^{6n + 2}}{1 + q^{6n + 2}} - \frac{q^{6n + 5}}{1 - q^{6n + 5}}\right) = \frac{\phi^{2}(q)f(q)}{f(-q, q^{2})}$$ The fraction on the right side is $$\begin{align} \frac{\phi^{2}(q)f(q)}{f(-q, q^{2})} &= \frac{\phi^{2}(q)(-q;-q)_{\infty}}{(q; -q^{3})_{\infty}(-q^{2};-q^{3})_{\infty}(-q^{3};-q^{3})_{\infty}}\notag\\ &= \frac{\phi^{2}(q)(-q;-q)_{\infty}(q^{3};-q^{3})_{\infty}}{(q;-q)_{\infty}(-q^{3};-q^{3})_{\infty}}\notag\\ &= \frac{\phi^{3}(q)}{\phi(q^{3})}\notag\end{align}$$ Finally we obtain $$\frac{\phi^{3}(q)}{\phi(q^{3})} = 1 + 6\sum_{n = 0}^{\infty}\left(\frac{q^{6n + 1}}{1 - q^{6n + 1}} - \frac{q^{6n + 4}}{1 + q^{6n + 4}} + \frac{q^{6n + 2}}{1 + q^{6n + 2}} - \frac{q^{6n + 5}}{1 - q^{6n + 5}}\right)\tag{7}$$ Replacing $q$ by $(-q)$ we get $$\frac{\phi^{3}(-q)}{\phi(-q^{3})} = 1 + 6\sum_{n = 0}^{\infty}\left(\frac{q^{3n + 2}}{1 + q^{3n + 2}} - \frac{q^{3n + 1}}{1 + q^{3n + 1}}\right)\tag{8}$$ From the equations $(6), (7), (8)$ we get $$2\cdot\frac{\psi^{3}(q)}{\psi(q^{3})} - \frac{\phi^{3}(q)}{\phi(q^{3})} = \frac{\phi^{3}(-q^{2})}{\phi(-q^{6})}\tag{9}$$ The above relation is a non-trivial, non-obvious connection between the theta functions of argument $q$ and $q^{3}$.

Transcription Formulas

Let us now transform the above relation in the form of a modular equation of degree $3$. Let $x, y, z$ correspond to $q$ in the following manner: $$\begin{align}x &= 1 - \frac{\phi^{4}(-q)}{\phi^{4}(q)}\notag\\ y &= \pi\cdot\dfrac{_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2}; 1; 1 - x\right)}{_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2}; 1; x\right)}\notag\\ z &= _{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2}; 1; x\right)\notag\\ q &= e^{-y}\notag\end{align}$$ Similarly let $x_{1}, y_{1}, z_{1}$ correspond to $q^{3}$. Clearly we can then see that $y_{1} = 3y$.

Then we have $\phi(q) = \sqrt{z}$ and $\phi(q^{3}) = \sqrt{z_{1}}$ and $$\begin{align}\phi(-q) &= \sqrt{\phi^{2}(-q)}\notag\\ &= \sqrt{\phi^{2}(q)\cdot\frac{\phi^{2}(-q)}{\phi^{2}(q)}}\notag\\ &= \sqrt{z\cdot\sqrt{1 - x}}\notag\\ &= \sqrt{z}(1 - x)^{1/4}\notag\end{align}$$ $$\begin{align}\phi(-q^{2}) &= \sqrt{\phi^{2}(-q^{2})}\notag\\ &= \sqrt{\phi(q)\phi(-q)}\notag\\ &= \sqrt{\sqrt{z}\cdot\sqrt{z}(1 - x)^{1/4}}\notag\\ &= \sqrt{z}(1 - x)^{1/8}\notag\end{align}$$ Finding $\psi(q)$ is bit involved. First we start with $\psi(q^{2})$ as follows: $$\begin{align}\psi(q^{2}) &= \frac{1}{2}q^{-1/4}\{\phi^{4}(q) - \phi^{4}(-q)\}^{1/4}\notag\\ &= \frac{1}{2}q^{-1/4}\{z^{2} - z^{2}(1 - x)\}^{1/4}\notag\\ &= \frac{1}{2}e^{y/4}\sqrt{z}x^{1/4}\notag\end{align}$$ and then $$\begin{align} \psi(q) &= \sqrt{\phi(q)\psi(q^{2})}\notag\\ &= \sqrt{\sqrt{z}\frac{1}{2}e^{y/4}\sqrt{z}x^{1/4}}\notag\\ &= \sqrt{\frac{z}{2}}(xe^{y})^{1/8}\notag\end{align}$$ Again $$\begin{align}\psi(-q) &= \psi(q)\sqrt{\frac{\phi(-q)}{\phi(q)}}\notag\\ &= \sqrt{\frac{z}{2}}e^{y/8}\{x(1 - x)\}^{1/8}\notag\\ \chi(q) &= \sqrt[3]{\frac{\phi(q)}{\psi(-q)}}\notag\\ &= \sqrt[3]{\dfrac{\sqrt{z}}{\sqrt{\dfrac{z}{2}}e^{y/8}\{x(1 - x)\}^{1/8}}}\notag\\ &= 2^{1/6}\{x(1 - x)e^{y}\}^{-1/24}\notag\\ \chi(-q) &= \frac{\chi(q)\psi(-q)}{\psi(q)}\notag\\ &= 2^{1/6}\{x(1 - x)e^{y}\}^{-1/24}(1 - x)^{1/8}\notag\\ &= 2^{1/6}(1 - x)^{1/12}(xe^{y})^{-1/24}\notag\\ \chi(-q^{2}) &= \chi(q)\chi(-q)\notag\\ &= 2^{1/3}(1 - x)^{1/24}(xe^{y})^{-1/12}\notag\\ f(q) &= \frac{\phi(q)}{\chi(q)}\notag\\ &= 2^{-1/6}\sqrt{z}\{x(1 - x)e^{y}\}^{1/24}\notag\\ f(-q) &= \sqrt[3]{\phi^{2}(-q)\psi(q)}\notag\\ &= 2^{-1/6}\sqrt{z}(1 - x)^{1/6}(xe^{y})^{1/24}\notag\\ f(-q^{2}) &= \sqrt[3]{\phi(-q)\psi^{2}(q)}\notag\\ &= 2^{-1/3}\sqrt{z}\{x(1 - x)e^{y}\}^{1/12}\notag\\ f(-q^{4}) &= \sqrt[3]{\phi(-q^{2})\psi^{2}(q^{2})}\notag\\ &= 4^{-1/3}\sqrt{z}(1 - x)^{1/24}(xe^{y})^{1/6}\notag\end{align}$$

Modular Equation of Degree 3

Putting the expressions of $\phi(q), \psi(q), \phi(-q^{2})$ in relation $(9)$ and doing the same for these functions when $q$ is replaced by $q^{3}$ we get $$2\cdot\frac{2^{-3/2}z^{3/2}(xe^{y})^{3/8}}{2^{-1/2}z_{1}^{1/2}(x_{1}e^{y_{1}})^{1/8}} - \frac{z^{3/2}}{z_{1}^{1/2}} = \frac{z^{3/2}(1 - x)^{3/8}}{z_{1}^{1/2}(1 - x_{1})^{1/8}}$$ Noting that $y_{1} = 3y$ the above equation simplifies to $$\left(\frac{x^{3}}{x_{1}}\right)^{1/8} - \left(\frac{(1 - x)^{3}}{1 - x_{1}}\right)^{1/8} = 1$$ In modern notation $x = k^{2}$ and $x_{1} = l^{2}$ where $L'/L = 3K'/K$, but Ramanujan used $\alpha$ for $x$ and $\beta$ for $x_{1}$ so that the above equation becomes $$\left(\frac{\alpha^{3}}{\beta}\right)^{1/8} - \left(\frac{(1 - \alpha)^{3}}{1 - \beta}\right)^{1/8} = 1$$ and we say that $\beta$ is of degree 3 over $\alpha$. Now $L'/L = 3K'/K$ implies that $K/K' = 3L/L'$ and hence we can see that $(1 -\alpha)$ is of degree 3 over $(1 - \beta)$. In general if $\beta$ is of degree $n$ over $\alpha$ then $(1 - \alpha)$ is of degree $n$ over $(1 - \beta)$. Thus in any modular equation containing only $\alpha$ and $\beta$ we can replace $\beta$ by $(1 - \alpha)$ and $\alpha$ by $(1 - \beta)$. Thus we finally obtain $$\left(\frac{\alpha^{3}}{\beta}\right)^{1/8} - \left(\frac{(1 - \alpha)^{3}}{1 - \beta}\right)^{1/8} = 1 = \left(\frac{(1 -\beta)^{3}}{1 - \alpha}\right)^{1/8} - \left(\frac{\beta^{3}}{\alpha}\right)^{1/8}\tag{10}$$ In the next post we will derive more such modular equations by first establishing the relationship between theta functions of $q$ and $q^{3}$ and then transcribing them.

Print/PDF Version