Elementary Approach to Modular Equations: Ramanujan's Theory 7

Continuing from previous post we proceed to derive further modular equations of degree $5$ in this post. Clearly in order to establish such equation we need to establish further theta function identities. This time we establish an identity concerning Ramanujan's $\psi$ function.

Identity Concerning $\psi(q)$ of Degree $5$

We will establish the following identity $$\psi^{2}(q^{2}) - q^{2}\psi^{2}(q^{10}) = \frac{\phi(-q^{10})f(-q^{10})}{\chi(-q^{2})}\tag{1}$$
To establish this we note that from equation $(4)$ of an earlier post we have $$\begin{align}\psi^{2}(q^{2}) - q^{2}\psi^{2}(q^{10}) &= \sum_{n = 0}^{\infty}\frac{q^{n}}{1 + q^{2n + 1}} - \sum_{n = 0}^{\infty}\frac{q^{5n + 2}}{1 + q^{10n + 5}}\notag\\ &= \frac{1}{1 + q} + \frac{q}{1 + q^{3}} + \frac{q^{3}}{1 + q^{7}} + \cdots\notag\end{align}$$ In the above expansion on RHS the exponents of $q$ in the numerator are all non-negative integers except those equal to $2$ modulo $5$ and hence we can write: $$\begin{align}&\psi^{2}(q^{2}) - q^{2}\psi^{2}(q^{10})\notag\\ &\,\,\,\,\,\,\,\,= \sum_{n = 1}^{\infty}\left(\frac{q^{5n - 1}}{1 + q^{10n - 1}} + \frac{q^{5n - 2}}{1 + q^{10n - 3}} + \frac{q^{5n - 4}}{1 + q^{10n - 7}} + \frac{q^{5n - 5}}{1 + q^{10n - 9}}\right)\tag{2}\end{align}$$ Next from equation $(3)$ of last post we have $$a\frac{f(-b/a, -a^{3}b)}{f(-a^{2}, -b^{2}}\phi(ab)\psi^{2}(a^{2}b^{2}) = \sum_{n = 1}^{\infty}\left(\frac{a^{n}b^{n - 1}}{1 - a^{2n}b^{2n - 2}} - \frac{a^{n - 1}b^{n}}{1 - a^{2n - 2}b^{2n}}\right)\tag{3}$$ In the above equation we put $a = iq^{1/2}, b = -iq^{9/2}$ and on simplification we get $$\phi(q^{5})\psi(q^{10})\frac{f(q^{4}, q^{6})}{f(q, q^{9})} = \sum_{n = 1}^{\infty}\left(\frac{q^{5n - 1}}{1 + q^{10n - 1}} + \frac{q^{5n - 5}}{1 + q^{10n - 9}}\right)\tag{4}$$ Again putting $a = iq^{3/2}, b = -iq^{7/2}$ in equation $(3)$ we get $$\phi(q^{5})\psi(q^{10})\frac{f(q^{2}, q^{8})}{f(q^{3}, q^{7})} = \sum_{n = 1}^{\infty}\left(\frac{q^{5n - 3}}{1 + q^{10n - 3}} + \frac{q^{5n - 5}}{1 + q^{10n - 7}}\right)$$ $$\Rightarrow \phi(q^{5})\psi(q^{10})\frac{qf(q^{2}, q^{8})}{f(q^{3}, q^{7})} = \sum_{n = 1}^{\infty}\left(\frac{q^{5n - 2}}{1 + q^{10n - 3}} + \frac{q^{5n - 4}}{1 + q^{10n - 7}}\right)\tag{5}$$ Adding equations $(4)$ and $(5)$ and using equation $(2)$ we get $$\psi^{2}(q^{2}) - q^{2}\psi^{2}(q^{10}) = \phi(q^{5})\psi(q^{10})\left\{\frac{f(q^{4}, q^{6})}{f(q, q^{9})} + q\frac{f(q^{2}, q^{8})}{f(q^{3}, q^{7})}\right\}\tag{6}$$ From the previous post we have $$\begin{align}f(a, b)f(c, d) + f(-a, -b)f(-c, -d) &= 2f(ac, bd)f(ad, bc)\notag\\ f(a, b)f(c, d) - f(-a, -b)f(-c, -d) &= 2af\left(\frac{b}{c}, \frac{c}{b}abcd\right)f\left(\frac{b}{d}, \frac{d}{b}abcd\right)\notag\end{align}$$ provided that $ab = cd$. Adding these two conditional identities we get $$f(a, b)f(c, d) = f(ac, bd)f(ad, bc) + af\left(\frac{b}{c}, \frac{c}{b}abcd\right)f\left(\frac{b}{d}, \frac{d}{b}abcd\right)$$ Puting $a = q, b = q^{4}, c = q^{2}, d = q^{3}$ in the above equation we get $$f(q, q^{4})f(q^{2}, q^{3}) = f(q^{3}, q^{7})f(q^{4}, q^{6}) + qf(q^{2}, q^{8})f(q, q^{9})\tag{7}$$ Using this identity in equation $(6)$ we get $$\psi^{2}(q^{2}) - q^{2}\psi^{2}(q^{10}) =\phi(q^{5})\psi(q^{10})\frac{f(q, q^{4})f(q^{2}, q^{3})}{f(q, q^{9})f(q^{3}, q^{7})}\tag{8}$$ Now we have $$\begin{align}f(q, q^{4})f(q^{2}, q^{3}) &= (-q;q^{5})_{\infty}(-q^{2};q^{5})_{\infty}(-q^{3};q^{5})_{\infty}(-q^{4};q^{5})_{\infty}f^{2}(-q^{5})\notag\\ &= \frac{(-q; q)_{\infty}}{(-q^{5};q^{5})_{\infty}}f^{2}(-q^{5})\notag\\ &= \frac{(q^{2}; q^{2})_{\infty}}{(q; q)_{\infty}}\frac{(q^{5}; q^{5})_{\infty}}{(q^{10}; q^{10})_{\infty}}f^{2}(-q^{5})\notag\\ &= \frac{(q^{5}; q^{10})_{\infty}}{(q; q^{2})_{\infty}}f^{2}(-q^{5})\notag\\ &= \frac{\chi(-q^{5})}{\chi(-q)}f^{2}(-q^{5})\notag\\ &= \frac{\phi(-q^{5})f(-q^{5})}{\chi(-q)}\notag\end{align}$$ and from previous post we have $$f(q, q^{9})f(q^{3}, q^{7}) = \chi(q)f(-q^{5})f(-q^{20})$$ Using these values in equation $(8)$ we get $$\begin{align}\psi^{2}(q^{2}) - q^{2}\psi^{2}(q^{10}) &= \phi(q^{5})\psi(q^{10})\frac{\phi(-q^{5})}{\chi(q)\chi(-q)f(-q^{20})}\notag\\ &= \frac{\phi^{2}(-q^{10})\psi(q^{10})}{\chi(-q^{2})f(-q^{20})}\notag\\ &= \frac{\phi(-q^{10})f(-q^{10})}{\chi(-q^{2})}\notag\end{align}$$ and thereby the starting equation $(1)$ is established. Most of the simplification has been done using properties of the functions $f,\phi, \psi, \chi$ established in an earlier post.

Modular Equation of Degree $5$

From equation $(13)$ of previous post we have $$\phi^{2}(q) - \phi^{2}(q^{5}) = 4q\chi(q)f(-q^{5})f(-q^{20})$$ Dividing equation $(1)$ by the above equation we get $$\frac{\psi^{2}(q^{2}) - q^{2}\psi^{2}(q^{10})}{\phi^{2}(q) - \phi^{2}(q^{5})} = \frac{\phi(-q^{10})f(-q^{10})}{4q\chi(q)\chi(-q^{2})f(-q^{5})f(-q^{20})}\tag{9}$$ It is now a simple matter of transcription to get the modular equation of degree $5$ from equation $(9)$. Using the transcription formulas established in an earlier post we get from equation $(9)$ $$\begin{align}&\dfrac{\dfrac{1}{4}e^{y/2}zx^{1/2} - e^{-2y}\dfrac{1}{4}e^{y_{1}/2}z_{1}x_{1}^{1/2}}{z - z_{1}}\notag\\ &\,\,\,\,\,\,\,\,= \dfrac{\dfrac{z_{1}^{1/2}(1 - x_{1})^{1/8}\cdot 2^{-1/3}z_{1}^{1/2}\{x_{1}(1 - x_{1})e^{y_{1}}\}^{1/12}}{2^{1/3}(1 - x)^{1/24}(xe^{y})^{-1/12}}}{2^{4/3}z_{1}\{x_{1}(1 - x_{1})\}^{5/24}\{x(1 - x)\}^{-1/24}}\notag\end{align}$$ Noting that $y_{1} = 5y$ and $x = \alpha, x_{1} = \beta$ and $m = z/z_{1}$ we can see that the above equation becomes $$\frac{m\sqrt{\alpha} - \sqrt{\beta}}{m - 1} = \left(\frac{\alpha}{\beta}\right)^{1/8}$$ $$\Rightarrow m\left\{\sqrt{\alpha} - \left(\frac{\alpha}{\beta}\right)^{1/8}\right\} = \sqrt{\beta} - \left(\frac{\alpha}{\beta}\right)^{1/8}$$ $$\Rightarrow m = \dfrac{\sqrt{\beta} - \left(\dfrac{\alpha}{\beta}\right)^{1/8}}{\sqrt{\alpha} - \left(\dfrac{\alpha}{\beta}\right)^{1/8}}$$ $$\Rightarrow m = \dfrac{1 - \left(\dfrac{\beta^{5}}{\alpha}\right)^{1/8}}{1 - (\alpha^{3}\beta)^{1/8}}\tag{10}$$ Changing $m$ into $5/m$, $\alpha$ into $(1 - \beta)$, $\beta$ into $(1 - \alpha)$ we get another modular equation $$\frac{5}{m} = \dfrac{1 - \left(\dfrac{(1 - \alpha)^{5}}{1 - \beta}\right)^{1/8}}{1 - \{(1 - \alpha)(1 -  \beta)^{3}\}^{1/8}}\tag{11}$$ Using the above modular equations it is possible to parametrize $\alpha, \beta$ in terms of $m$ (as was done in case of modular equations of degree $3$). With the parametrization in place it is possible to verify many more modular equations given by Ramanujan. We start with $$u = \left(\frac{\alpha^{5}}{\beta}\right)^{1/24},\,\, v = \left(\frac{\beta^{5}}{\alpha}\right)^{1/24}$$ It is clear that $uv^{2} = (\alpha\beta^{3})^{1/8}, u^{2}v = (\alpha^{3}\beta)^{1/8}$ and hence the equation $(10)$ can be written as $$\begin{align} m &= \frac{1 - v^{3}}{1 - u^{2}v}\notag\\ \Rightarrow m - 1 &= v(mu^{2} - v^{2})\tag{12}\end{align}$$ Similarly the equation $(17)$ of last post can be written as $$\begin{align}\frac{5}{m} &= \frac{1 + u^{3}}{1 + uv^{2}}\notag\\ \Rightarrow v^{2} &= \frac{m(1 + u^{3}) - 5}{5u}\tag{13}\end{align}$$ Again on squaring $(12)$ we get $$(m - 1)^{2} = v^{2}(mu^{2} - v^{2})^{2}$$ Substituting value of $v^{2}$ from $(13)$ we get $$(m - 1)^{2} = \left(\frac{m(1 + u^{3}) - 5}{5u}\right)\left(mu^{2} - \frac{m(1 + u^{3}) - 5}{5u}\right)^{2}$$ $$\Rightarrow 125u^{3}(m - 1)^{2} = \{m + mu^{3} - 5\}\{4mu^{3} - m + 5\}^{2}$$ Putting $u^{3} = t$ we get $$125t(m - 1)^{2} = \{m + mt - 5\}\{4mt - m + 5\}^{2}$$ Note that $t = -1$ satisfies this equation identically and thus on cancelling the factor $(t + 1)$ after simplification we get $$16m^{3}t^{2} - (8m^{3} + 40m^{2})t + m^{3} - 15m^{2} + 75m - 125 = 0$$ $$\Rightarrow 16m^{3}t^{2} - 8m^{2}(m + 5)t + (m - 5)^{3} = 0$$ $$\begin{align}\Rightarrow t &= \frac{8m^{2}(m + 5) \pm \sqrt{64m^{4}(m + 5)^{2} - 64m^{3}(m - 5)^{3}}}{32m^{3}}\notag\\ &= \frac{m^{2} + 5m \pm \sqrt{m^{2}(m + 5)^{2} - m(m - 5)^{3}}}{4m^{2}}\notag\\ &= \frac{m^{2} + 5m \pm 5\sqrt{m^{3} - 2m^{2} + 5m}}{4m^{2}}\notag\\ &= \frac{m^{2} + 5m \pm 5\rho}{4m^{2}}\notag\end{align}$$ where $\rho$ denotes the quantity $\sqrt{m^{3} - 2m^{2} + 5m}$ (following Bruce C. Berndt).

Now as $q \to 0+$ we see that $m \to \phi^{2}(0)/\phi^{2}(0) = 1$ and $v \to 0$. Thus from the relation $5/m = (1 + u^{3})/(1 + uv^{2})$ we see that $t = u^{3} \to 4$ as $q \to 0+$. Hence in the expression for $t$ above we must take the $+$ sign before the square root. Then we have $$u^{3} = \frac{m^{2} + 5m + 5\rho}{4m^{2}}$$ In a similar manner it is possible to eliminate $u$ from the relations connecting $u, v, m$ and get expression for $v^{3}$ in terms of $m$.

However Berndt follows a very simple approach based on symmetry of relations connecting $u, v, m$. When $u, v, m$ are replaced by $-v, -u, 5/m$ then the relations between $u, v, m$ remain invariant and hence we can apply these replacements in the expression for $u^{3}$ obtained above. Doing so leads to the following expression for $v^{3}$: $$v^{3} = \frac{\rho - m - 1}{4}$$ Thus we finally obtain $$\left(\frac{\alpha^{5}}{\beta}\right)^{1/8} = \frac{5\rho + m^{2} + 5m}{4m^{2}},\, \left(\frac{\beta^{5}}{\alpha}\right)^{1/8} = \frac{\rho - m - 1}{4}\tag{14}$$ where $$\rho = \sqrt{m^{3} - 2m^{2} + 5m}$$ Using the above parametrization many new modular equations can be obtained. It is not clear whether Ramanujan derived his modular equations in this fashion or he used some theta function identities to derive them. We will establish one more modular equation in order to show the usefulness of these parametrizations in terms of $m$. For more such equations the reader should consult Ramanujan's Notebooks Part III by Bruce C. Berndt.

From the definitions of $u, v$ and $(14)$ we can see that $$\begin{align}(\alpha^{3}\beta)^{1/8} &= u^{2}v\notag\\ &= 1 - \frac{1 - v^{3}}{m}\notag\\ &= \frac{v^{3} + m - 1}{m}\notag\\ &= \dfrac{\dfrac{\rho - m - 1}{4} + m - 1}{m}\notag\\ &= \frac{\rho + 3m - 5}{4m}\notag\end{align}$$ and $$\begin{align}(\alpha\beta^{3})^{1/8} &= uv^{2}\notag\\ &= \frac{m}{5}(1 + u^{3}) - 1\notag\\ &= \frac{m}{5}\left(1 + \frac{m^{2} + 5m + 5\rho}{4m^{2}}\right) - 1\notag\\ &= \frac{m^{2} + m + \rho}{4m} - 1\notag\\ &= \frac{\rho + m^{2} - 3m}{4m}\notag\end{align}$$ And then $$\begin{align}\left(\frac{\beta}{\alpha}\right)^{1/4} &= \frac{(\alpha\beta^{3})^{1/8}}{(\alpha^{3}\beta)^{1/8}}\notag\\ &= \frac{\rho + m^{2} - 3m}{\rho + 3m - 5}\notag\\ &= \frac{\rho + m^{2} - 3m}{\rho + 3m - 5}\frac{\rho - 3m + 5}{\rho - 3m + 5}\notag\\ &= \frac{(\rho + m^{2} - 3m)(\rho - 3m + 5)}{\rho^{2} - (3m - 5)^{2}}\notag\\ &= \frac{\rho^{2} + (m^{2} - 6m + 5)\rho - 3m^{3} + 14m^{2} - 15m}{\rho^{2} - 9m^{2} + 30m - 25}\notag\\ &= \frac{(m^{2} - 6m + 5)\rho - 2m(m^{2} - 6m + 5)}{m^{3} - 11m^{2} + 35m - 25}\notag\\ &= \frac{(m - 1)(m - 5)(\rho - 2m)}{(m - 1)(m - 5)^{2}}\notag\\ &= \frac{\rho - 2m}{m - 5}\tag{15}\end{align}$$ Inverting we get $$\begin{align}\left(\frac{\alpha}{\beta}\right)^{1/4} &= \frac{m - 5}{\rho - 2m}\notag\\ &= \frac{m - 5}{\rho - 2m}\frac{\rho + 2m}{\rho + 2m}\notag\\ &= \frac{(m - 5)(\rho + 2m)}{m(m - 1)(m - 5)}\notag\\ &= \frac{\rho + 2m}{m(m - 1)}\tag{16}\end{align}$$ Replacing $m$ by $5/m$, $\alpha$ by $(1 - \beta)$, $\beta$ by $(1 - \alpha)$ in $(16)$ we get $$\begin{align}\left(\frac{1 - \beta}{1 - \alpha}\right)^{1/4} &= \dfrac{\sqrt{\dfrac{125}{m^{3}} - \dfrac{50}{m^{2}} + \dfrac{25}{m}} + \dfrac{10}{m}}{\dfrac{5}{m}\left(\dfrac{5}{m} - 1\right)}\notag\\ &= \frac{\rho + 2m}{5 - m}\tag{17}\end{align}$$ Multiplying $(15)$ and $(17)$ we get $$\begin{align}\left(\frac{\beta(1 - \beta)}{\alpha(1 - \alpha)}\right)^{1/4} &= \frac{\rho - 2m}{m - 5}\frac{\rho + 2m}{5 - m}\notag\\ &= \frac{m(m - 1)(m - 5)}{(m - 5)(5 - m)}\notag\\ &= \frac{m(m - 1)}{5 - m}\tag{18}\end{align}$$ Adding $(15)$ and $(17)$ and using $(18)$ we get $$\begin{align}\left(\frac{\beta}{\alpha}\right)^{1/4} + \left(\frac{1 - \beta}{1 - \alpha}\right)^{1/4} &= \frac{4m}{5 - m}\notag\\ &= \frac{5m - m}{5 - m} = \frac{5m - m^{2} + m(m - 1)}{5 - m}\notag\\ &= m + \frac{m(m - 1)}{5 - m}\notag\\ &= m + \left(\frac{\beta(1 - \beta)}{\alpha(1 - \alpha)}\right)^{1/4}\notag\end{align}$$ And then we have finally in the style of Ramanujan $$m = \left(\frac{\beta}{\alpha}\right)^{1/4} + \left(\frac{1 - \beta}{1 - \alpha}\right)^{1/4} - \left(\frac{\beta(1 - \beta)}{\alpha(1 - \alpha)}\right)^{1/4}\tag{19}$$ and its reciprocal equation $$\frac{5}{m} = \left(\frac{\alpha}{\beta}\right)^{1/4} + \left(\frac{1 - \alpha}{1 - \beta}\right)^{1/4} - \left(\frac{\alpha(1 - \alpha)}{\beta(1 - \beta)}\right)^{1/4}\tag{20}$$ It turns out that obtaining modular equations using theta function identities (and subsequent algebraic manipulations) requires great ingenuity and there does not seem to be any general method by which such identities can be obtained. And with increasing degree, the identities and the resulting modular equations become quite complex. In this area of mathematics no one has matched the genius of Ramanujan and his innate sense of finding such identities.

The modern approach (e.g. followed in Bruce C. Berndt's books for higher degree modular equations) is based on the theory of modular forms and it can only be used to prove / disprove a given identity. It does not lead to a way of finding such identities. Often the method involves heavy calculations which are normally performed using some computer algebra packages. On the theoretical side the theory of modular forms itself requires heavy machinery. To this day the approach and methods of Ramanujan remain a mystery. In my opinion his methods must have been elementary in nature and highly economical in the sense that the form of Ramanujan's modular equations were far simpler than that provided by other mathematicians.

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