# Ramanujan's Class Invariants

After a heavy discussion on the modular equations found by Ramanujan, we will now focus on another significant discovery of his namely "Class Invariants".

### Class Invariant

A modular equation is a relation between two moduli $k \in (0, 1)$ and $l \in (0, 1)$, such that the complete elliptic integrals for these moduli have the following relation: $$\frac{L'}{L} = n\cdot\frac{K'}{K}$$ Here $n$ is a positive rational number. From the past series of posts we know that the relation between $k, l$ is an algebraic one, meaning there is a polynomial $P(x, y)$ with rational coefficients such that $P(k, l) = 0$. Often the modular equation is not expressed as a polynomial but rather as an equation involving fractional powers of $k, l$. Ramanujan expressed his equations in terms of $\alpha = k^{2}$ and $\beta = l^{2}$.

Ramanujan used the modular equation in a very clever way and put a further condition on the moduli $k, l$ such that $l = k'$ in which case $L = K'$ and $L' = K$. Then we arrive at the relation: $$\frac{L'}{L} = \sqrt{n},\,\,\,\,\frac{K'}{K} = \frac{1}{\sqrt{n}}$$ and $P(k, k') = 0$ i.e. $P(k, \sqrt{1 - k^{2}}) = 0$ so that $k, l = k' = \sqrt{1 - k^{2}}$ are both algebraic numbers. Thus if $q = e^{-\pi\sqrt{n}}$ then the corresponding modulus $k = k(q) = \theta_{2}^{2}(q)/\theta_{3}^{2}(q)$ is an algebraic number. Such moduli are called singular moduli. Clearly from the above it is also seen that $k' = k(e^{-\pi/\sqrt{n}})$.

A direct evaluation of the singular moduli is not done by using the equation $P(k, k') = 0$, but rather Ramanujan introduced two functions $G_{n}, g_{n}$ related with the modulus $k = k(q) = k(e^{-\pi\sqrt{n}})$ as follows: \begin{align}G_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 + e^{-\pi\sqrt{n}})(1 + e^{-3\pi\sqrt{n}})(1 + e^{-5\pi\sqrt{n}})\cdots\tag{1}\\ g_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 - e^{-\pi\sqrt{n}})(1 - e^{-3\pi\sqrt{n}})(1 - e^{-5\pi\sqrt{n}})\cdots\tag{2}\end{align} i.e. \begin{align} G_{n} &= 2^{-1/4}q^{-1/24}(1 + q)(1 + q^{3})(1 + q^{5})\cdots\notag\\ &= 2^{-1/4}q^{-1/24}(-q;q^{2})_{\infty}\notag\\ &= 2^{-1/4}q^{-1/24}\chi(q)\notag\\ &= G(q)\tag{3}\\ g_{n} &= 2^{-1/4}q^{-1/24}(1 - q)(1 - q^{3})(1 - q^{5})\cdots\notag\\ &= 2^{-1/4}q^{-1/24}(q;q^{2})_{\infty}\notag\\ &= 2^{-1/4}q^{-1/24}\chi(-q)\notag\\ &= g(q)\tag{4}\end{align} From the expressions of $k$ and $k'$ in terms of $q$ we see that $$G_{n} = (2kk')^{-1/12},\,\,\,\, g_{n} = \left(\frac{2k}{k'^{2}}\right)^{-1/12}$$ where $k = k(q) = k(e^{-\pi\sqrt{n}})$. It is easily seen that $G_{n}, g_{n}$ are algebraic functions of $k$ and that given the value of $G_{n}, g_{n}$ the value of $k$ is obtained by the solution of a quadratic equation whose coefficients are rational functions of $G_{n}, g_{n}$.

It turns out that in practice finding the value of $G_{n}, g_{n}$ is somewhat easier than finding the value of $k$. Ramanujan called these functions $G_{n}, g_{n}$ as class invariants. From the definition we can see that if we replace $n$ by $1/n$ then $k$ changes to $k'$ and vice-versa, so that the product $kk'$ remains unchanged. Thus we have $G_{1/n} = G_{n}$

Again if we replace $n$ by $4n$, $q$ changes to $q^{2}$ and hence $g_{4n} = g(q^{2})$. And we have \begin{align} G_{n}g_{n} &= 2^{-1/2}q^{-1/12}(1 - q^{2})(1 - q^{6})(1 - q^{10})\cdots\notag\\ &= 2^{-1/2}q^{-1/12}\cdot 2^{1/4}q^{1/12}g(q^{2})\notag\\ &= 2^{-1/4}g_{4n}\notag\end{align} so that $g_{4n} = 2^{1/4}g_{n}G_{n}$.

Again we can see that \begin{align} g_{4/n} &= 2^{1/4}g_{1/n}G_{1/n}\notag\\ &= 2^{1/4}g_{1/n}G_{n}\notag\\ &= 2^{1/4}\left(\frac{2k'}{k^{2}}\right)^{-1/12}(2kk')^{-1/12}\notag\\ &= 8^{1/12}\left(\frac{k^{2}}{2k'}\cdot\frac{1}{2kk'}\right)^{1/12}\notag\\ &= \left(\frac{2k}{k'^{2}}\right)^{1/12} = 1/g_{n}\notag\end{align} Similarly using the expressions of $G_{n}, g_{n}$ in the form of $k$ we can prove that $(g_{n}G_{n})^{8}(G_{n}^{8} - g_{n}^{8}) = 1/4$. We can summarize the elementary properties of the class invariants as follows: $$g_{4n} = 2^{1/4}g_{n}G_{n}\tag{5}$$ $$G_{1/n} = G_{n},\,\,\,\, 1/g_{n} = g_{4/n}\tag{6}$$ $$(g_{n}G_{n})^{8}(G_{n}^{8} - g_{n}^{8}) = \frac{1}{4}\tag{7}$$ The calculation of $G_{n}, g_{n}$ often requires the use of modular equations. For example if $n = 3$ the modular equation is given by $$\sqrt{kl} + \sqrt{k'l'} = 1$$ In the above if we put $k' = l$ we get $\sqrt{kk'} = 1/2$ or $(2kk') = 1/2$ and hence $G_{3} = 2^{1/12}$.

To calculate $G_{5}$ we need to use a modular equation of degree $5$. Clearly we can use the equation $$m = 1 + 2^{4/3}\left(\frac{\beta^{5}(1 - \beta)^{5}}{\alpha(1 - \alpha)}\right)^{1/24}$$ and its cousin $$\frac{5}{m} = 1 + 2^{4/3}\left(\frac{\alpha^{5}(1 - \alpha)^{5}}{\beta(1 - \beta)}\right)^{1/24}$$ For the calculation of $G_{5}$ we apply the restriction $\alpha + \beta = 1$ and then $G_{5} = 2^{-1/12}(\alpha\beta)^{-1/24}$. We then have $$m = 1 + 2^{4/3}(\alpha\beta)^{1/6},\,\,\,\, \frac{5}{m} = 1 + 2^{4/3}(\alpha\beta)^{1/6}$$ so that by multiplication we get \begin{align}&5 = (1 + 2^{4/3}(\alpha\beta)^{1/6})^{2}\notag\\ &\Rightarrow 2^{1/3}(\alpha\beta)^{1/6} = \frac{\sqrt{5} - 1}{2}\notag\\ &\Rightarrow 2^{-1/3}(\alpha\beta)^{-1/6} = \frac{2}{\sqrt{5} - 1}\notag\\ &\Rightarrow G_{5} = 2^{-1/12}(\alpha\beta)^{-1/24} = \left(\frac{\sqrt{5} + 1}{2}\right)^{1/4}\tag{8}\end{align}

### Modular Equation in Schlafli Form

Often a modular equation may be given in the form of a relation between $G(q)$ and $G(q^{n})$ (correspondingly as a relation between $g(q)$ and $g(q^{n})$). Such modular equation are also called modular equations of Schlafli type. Ramanujan tells us that in these cases if $n$ is odd, then the same equation can be used to calculate the value of $G_{n}$ and $g_{2n}$. For let $P(G(q^{n}), G(q)) = 0$ be the given relation. If we put $q = e^{-\pi/\sqrt{n}}$ then we have $$u = G(q^{n}) = G_{n}, v = G(q) = G_{1/n} = G_{n} = u$$ and thus the value of $u = G_{n}$ is found by solving $P(u, u) = 0$.

If on the other hand we have the relation $P(g(q^{n}), g(q)) = 0$ then we set $q = e^{-\pi\sqrt{2/n}}$ so that $$v = g(q) = g_{2/n}, u = g(q^{n}) = g_{2n} = 1/g_{2/n} = 1/v$$ and then $u = g_{2n}$ is found by solving $P(u, 1/u) = 0$.

In his paper "Modular Equations and Approximations to $\pi$" Ramanujan applies this technique to evaluate $g_{10}$. The modular equation he uses is the following: $$\left(\frac{v}{u}\right)^{3} - \left(\frac{u}{v}\right)^{3} = 2\left(u^{2}v^{2} + \frac{1}{u^{2}v^{2}}\right)$$ where $u = g(q) = 2^{-1/4}q^{-1/24}\chi(q)$ and $v = g(q^{5})$.

To obtain this equation we use the modular equation of degree $5$ as follows: $$\left(1 + 2^{4/3}\left(\frac{\beta^{5}(1 - \beta)^{5}}{\alpha(1 - \alpha)}\right)^{1/24}\right)\left(1 + 2^{4/3}\left(\frac{\alpha^{5}(1 - \alpha)^{5}}{\beta(1 - \beta)}\right)^{1/24}\right) = 5$$ \begin{align}\Rightarrow 2^{4/3}\left(\frac{\beta^{5}(1 - \beta)^{5}}{\alpha(1 - \alpha)}\right)^{1/24} &+\, 2^{4/3}\left(\frac{\alpha^{5}(1 - \alpha)^{5}}{\beta(1 - \beta)}\right)^{1/24}\notag\\ &+\, 4\cdot 2^{2/3}\{\alpha\beta(1 - \alpha)(1 - \beta)\}^{1/6} = 4\notag\end{align} Noting that $\alpha(1 - \alpha) = G^{-24}(q)/4, \beta(1 - \beta) = G^{-24}(q^{5})/4$ we see that the above relation implies: $$2\left(\frac{G(q)}{G^{5}(q^{5})} + \frac{G(q^{5})}{G^{5}(q)}\right) + \frac{4}{G^{4}(q)G^{4}(q^{5})} = 4$$ Multiplying by $G^{2}(q)G^{2}(q^{5})$ we get $$\left(\frac{G(q)}{G(q^{5})}\right)^{3} + \left(\frac{G(q^{5})}{G(q)}\right)^{3} = 2 \left(G^{2}(q)G^{2}(q^{5}) - \frac{1}{G^{2}(q)G^{2}(q^{5})}\right)\tag{9}$$ Switching to the $\chi(-q)$ function we get $$\sqrt{q}\cdot\frac{\chi^{3}(-q)}{\chi^{3}(-q^{5})} + \frac{1}{\sqrt{q}}\cdot\frac{\chi^{3}(-q^{5})}{\chi^{3}(-q)} = 2\left(\frac{\chi^{2}(-q)\chi^{2}(-q^{5})}{2\sqrt{q}} - \frac{2\sqrt{q}}{\chi^{2}(-q)\chi^{2}(-q^{5})}\right)$$ Multiplying by $\sqrt{q}$ we get $$q\cdot\frac{\chi^{3}(-q)}{\chi^{3}(-q^{5})} + \frac{\chi^{3}(-q^{5})}{\chi^{3}(-q)} = \chi^{2}(-q)\chi^{2}(-q^{5}) - \frac{4q}{\chi^{2}(-q)\chi^{2}(-q^{5})}$$ Replacing $q$ by $(-q)$ and further dividing by $\sqrt{q}$ we get $$\frac{1}{\sqrt{q}}\cdot\frac{\chi^{3}(q^{5})}{\chi^{3}(q)} - \sqrt{q}\cdot\frac{\chi^{3}(q)}{\chi^{3}(q^{5})} = 2\left(\frac{\chi^{2}(q)\chi^{2}(q^{5})}{2\sqrt{q}} + \frac{2\sqrt{q}}{\chi^{2}(q)\chi^{2}(q^{5})}\right)$$ $$\Rightarrow \left(\frac{g(q^{5})}{g(q)}\right)^{3} - \left(\frac{g(q)}{g(q^{5})}\right)^{3} = 2\left(g^{2}(q)g^{2}(q^{5}) + \frac{1}{g^{2}(q)g^{2}(q^{5})}\right)\tag{10}$$ so that we get the desired relation connecting $u = g(q)$ and $v = g(q^{5})$ as: $$\left(\frac{v}{u}\right)^{3} - \left(\frac{u}{v}\right)^{3} = 2\left(u^{2}v^{2} + \frac{1}{u^{2}v^{2}}\right)$$ If $u = g_{2/5}$ then $v = g_{10} = 1/g_{2/5} = 1/u$ and hence we get $$v^{6} - v^{-6} = 4 \Rightarrow v^{6} = 2 + \sqrt{5} \Rightarrow v^{2} = \frac{1 + \sqrt{5}}{2}$$ Finally we get $$\displaystyle g_{10} = v = \sqrt{\frac{1 + \sqrt{5}}{2}}\tag{11}$$ Ramanujan calculated a host of class invariants mostly using modular equations. In his classic paper on modular equations he gave a list of all the class invariants he calculated. Some of these invariants were already obtained by H. Weber in a different context using different methods. Ramanujan's main reason for evaluating these values was that he used them in finding approximations to $\pi$. Also on a lighter note Ramanujan loved manipulating expressions containing radicals and calculation of these class invariants gave him enough radicals to play with.

As another example let us put $q = e^{-\pi}$ in equation $(9)$ and then $G(q) = G_{1} = 1$ and $G(q^{5}) = G_{25}$ and we have the following relation satisfied by $x = G_{25}$: $$x^{3} + \frac{1}{x^{3}} = 2\left(x^{2} - \frac{1}{x^{2}}\right)$$ $$\Rightarrow \left(x + \frac{1}{x}\right)\left(x^{2} + \frac{1}{x^{2}} - 1\right) = 2\left(x + \frac{1}{x}\right)\left(x - \frac{1}{x}\right)$$ Since $(x + (1/x)) \neq 0$ we have $$x^{2} + \frac{1}{x^{2}} - 1 = 2\left(x - \frac{1}{x}\right)$$ Putting $t = x - (1/x)$ we get $t^{2} - 2t + 1 = 0$ so that $t = 1$ and therefore \begin{align}&x - \frac{1}{x} = 1\notag\\ &\Rightarrow x^{2} - x - 1 = 0\notag\\ &\Rightarrow x = \frac{1 \pm \sqrt{5}}{2}\notag\end{align} Clearly $x = G_{25} > 0$ so we finally obtain $$G_{25} = \frac{1 + \sqrt{5}}{2}\tag{12}$$ Putting $y = g_{25}^{8}, a = G_{25}^{8}$ and using equation $(7)$ we get $$ay(a - y) = \frac{1}{4}$$ $$\Rightarrow 4ay^{2} - 4a^{2}y + 1 = 0$$ Since $y = g_{25}^{8} > 0$ it follows that \begin{align}y &= \frac{a^{2} + \sqrt{a^{4} - a}}{2a}\notag\\ &= \frac{G_{25}^{8} + G_{25}^{-8}\sqrt{G_{25}^{32} - G_{25}^{8}}}{2}\notag\\ &= \frac{G_{25}^{8} + G_{25}^{2}\sqrt{G_{25}^{12} - G_{25}^{-12}}}{2}\notag\\ &= \frac{G_{25}^{2}}{2}\left(G_{25}^{6} + \sqrt{G_{25}^{12} - G_{25}^{-12}}\right)\notag\\ &= \frac{G_{25}^{2}}{4}\left(2G_{25}^{6} + 2\sqrt{G_{25}^{12} - G_{25}^{-12}}\right)\notag\\ &= \frac{G_{25}^{2}}{4}\left(\sqrt{G_{25}^{6} + G_{25}^{-6}} + \sqrt{G_{25}^{6} - G_{25}^{-6}}\right)^{2}\notag\end{align} Now from $(12)$ we get $$G_{25}^{6} = 9 + 4\sqrt{5},\, G_{25}^{-6} = 9 - 4\sqrt{5}$$ and therefore we have \begin{align} g_{25}^{8} &= y = \frac{1}{4}\left(\frac{1 + \sqrt{5}}{2}\right)^{2}\left(\sqrt{18} + \sqrt{8\sqrt{5}}\right)^{2}\notag\\ &= \frac{1}{2}\left(\frac{1 + \sqrt{5}}{2}\right)^{2}\left(3 + 2\sqrt[4]{5}\right)^{2}\notag\end{align} Thus we get $$g_{25} = 2^{-3/8}(1 + \sqrt{5})^{1/4}(3 + 2\sqrt[4]{5})^{1/4}\tag{13}$$ and then from equation $(5)$ we get \begin{align}g_{100} &= 2^{1/4}g_{25}G_{25}\notag\\ &= 2^{-1/8}(1 + \sqrt{5})^{1/4}(3 + 2\sqrt[4]{5})^{1/4}\cdot\frac{1 + \sqrt{5}}{2}\notag\\ &= \frac{(1 + \sqrt{5})^{5/4}(3 + 2\sqrt[4]{5})^{1/4}}{2\sqrt[8]{2}}\tag{14}\end{align} We are not going to derive any further values of the class invariants for the simple reason that we have not derived the corresponding modular equations. However before concluding the post we would like to prove a formula of Ramanujan which helped him a lot in calculating class invariants. The formula is a relationship between the values of $g_{9n}$ and $g_{n}$. Later a similar counterpart was found by Borwein brothers (of "Pi and the AGM" fame) for the invariants $G_{9n}$ and $G_{n}$.

Theorem: If $p = G_{n}^{4} + G_{n}^{-4}$ then \begin{align} \frac{G_{9n}}{G_{n}} &= (p + \sqrt{p^{2} - 1})^{1/6}\left\{\sqrt{\frac{p^{2} - 2 + \sqrt{(p^{2} - 1)(p^{2} - 4)}}{2}}\right.\notag\\ &\left.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, +\, \sqrt{\frac{p^{2} - 4 + \sqrt{(p^{2} - 1)(p^{2} - 4)}}{2}}\right\}^{1/3}\tag{15}\end{align} The theorem by Ramanujan for $g_{n}, g_{9n}$ is as follows:
Theorem: If $p = g_{n}^{4} - g_{n}^{-4}$ then \begin{align} \frac{g_{9n}}{g_{n}} &= (p + \sqrt{p^{2} + 1})^{1/6}\left\{\sqrt{\frac{p^{2} + 2 + \sqrt{(p^{2} + 1)(p^{2} + 4)}}{2}}\right.\notag\\ &\left.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, +\, \sqrt{\frac{p^{2} + 4 + \sqrt{(p^{2} + 1)(p^{2} + 4)}}{2}}\right\}^{1/3}\tag{16}\end{align} The proof is based on a modular equation of degree 3 in Schlafli form. We first derive this modular equation. We start with the following modular equations from last post: $$\left(\frac{(1 -\beta)^{3}}{1 - \alpha}\right)^{1/8} - \left(\frac{\beta^{3}}{\alpha}\right)^{1/8} = 1$$ $$m = 1 + 2\left(\frac{\beta^{3}}{\alpha}\right)^{1/8}$$ Combining these we get $$m = \left(\frac{(1 -\beta)^{3}}{1 - \alpha}\right)^{1/8} + \left(\frac{\beta^{3}}{\alpha}\right)^{1/8}$$ and hence $$m^{2} - 1 = m^{2} - 1^{2} = 4\left(\frac{(1 -\beta)^{3}}{1 - \alpha}\right)^{1/8}\left(\frac{\beta^{3}}{\alpha}\right)^{1/8}$$ i.e. $$m^{2} = 1 + 4\left(\frac{\beta^{3}(1 - \beta)^{3}}{\alpha(1 - \alpha)}\right)^{1/8}$$ and therefore replacing $\alpha$ by $1 - \beta$ and $m$ by $3/m$ we get $$\frac{9}{m^{2}} = 1 + 4\left(\frac{\alpha^{3}(1 - \alpha)^{3}}{\beta(1 - \beta)}\right)^{1/8}$$ Multiplying these equations and simplifying we get $$\left(\frac{\beta^{3}(1 - \beta)^{3}}{\alpha(1 - \alpha)}\right)^{1/8} + \left(\frac{\alpha^{3}(1 - \alpha)^{3}}{\beta(1 - \beta)}\right)^{1/8} + 4\{\alpha\beta(1 - \alpha)(1 - \beta)\}^{1/4} = 2$$ i.e. \begin{align}\{\alpha(1 - \alpha)\}^{1/2} &+ \{\beta(1 - \beta)\}^{1/2}\notag\\ &+ 4\{\alpha\beta(1 - \alpha)(1 - \beta)\}^{3/8} = 2\{\alpha\beta(1 - \alpha)(1 - \beta)\}^{1/8}\notag\\ \Rightarrow \left(\frac{\beta(1 - \beta)}{\alpha(1 - \alpha)}\right)^{1/4} &+ \left(\frac{\alpha(1 - \alpha)}{\beta(1 - \beta)}\right)^{1/4}\notag\\ &+ 4\{\alpha\beta(1 - \alpha)(1 - \beta)\}^{1/8} = 2\{\alpha\beta(1 - \alpha)(1 - \beta)\}^{-1/8}\notag\end{align} This reduces to $(Q + Q^{-1}) + 2\sqrt{2}(P - P^{-1}) = 0$ where $$P = \{16\alpha\beta(1 - \alpha)(1 - \beta)\}^{1/8},\, Q = \left(\frac{\beta(1 - \beta)}{\alpha(1 - \alpha)}\right)^{1/4}$$ If $\alpha$ corresponds to $q = e^{-\pi\sqrt{n}}$ then $\beta$ corresponds $q^{3} = e^{-\pi\sqrt{9n}}$ and therefore we have $$G_{n} = \{4\alpha(1 - \alpha)\}^{-1/24}, G_{9n} = \{4\beta(1 - \beta)\}^{-1/24}$$ so that $P = (G_{n}G_{9n})^{-3}, Q = (G_{n}/G_{9n})^{6}$ and our modular equation becomes $$(G_{n}/G_{9n})^{6} + (G_{n}/G_{9n})^{-6} + 2\sqrt{2}\{(G_{n}G_{9n})^{-3} - (G_{n}G_{9n})^{3}\} = 0$$ Let $x = (G_{9n}/G_{n})^{3}$ and then we have the following equation $$x^{4} - 2\sqrt{2}G_{n}^{6}x^{3} + 2\sqrt{2}G_{n}^{-6}x + 1 = 0$$ Ramanujan is directly able to solve this equation by using the notation $p = G_{n}^{4} + G_{n}^{-4}$. After some reasonable amount of algebra it can be seen that the equation can be re-written as: $$(x^{2} - \sqrt{2}G_{n}^{6}x + p)^{2} = 2(p^{2} - 1)\left(G_{n}^{2}x - \frac{1}{\sqrt{2}}\right)^{2}$$ Numerical consideration (like $x > 1$) force us to take the positive square root on both sides and this leads to $$x^{2} - \sqrt{2}G_{n}^{6}x + p = \sqrt{2(p^{2} - 1)}\left(G_{n}^{2}x - \frac{1}{\sqrt{2}}\right)$$ so that $$x^{2} -\sqrt{2}G_{n}^{2}(G_{n}^{4} + \sqrt{p^{2} - 1})x + p + \sqrt{p^{2} - 1} = 0$$ Again we need to take the positive root and after a tedious algebra we arrive at the value of $x = (G_{9n}/G_{n})^{3}$ which matches with the value expected by the expression given in the theorem. The readers should verify this calculation in their leisure time. As a simple application if we use the fact that $G_{1} = 1$ then the above formula gives $$G_{9} = (2 + \sqrt{3})^{1/6} = \sqrt[3]{\frac{1 + \sqrt{3}}{\sqrt{2}}}$$ The relation between $g_{n}$ and $g_{9n}$ can be similarly proved. The idea is to use the existing modular equation in terms of $G_{n}$ and switch to $\chi(-q)$ function and after some simplification we can change sign of $q$ to get equation in the form of $\chi(q)$ which can be transformed to get an equation connecting $g_{n}, g_{9n}$. As an application of this formula $g_{2} = 1$ and hence $p = 0$ so that $g_{18} = (5 + 2\sqrt{6})^{1/6}$.

Using modular equations (given by Ramanujan and others) and using the above theorem a host of class invariants can be calculated, but there are some in Ramanujan's list which have not yet been calculated using these methods. Possibly Ramanujan used modular equations, but somehow failed to record those equations in his notebooks or he used some results similar to those theorems above (e.g. a relation between $g_{n}, g_{5n}$). We don't know the routes he took to find all these class invariants. Modern authors have used the theory of modular forms and Kronecker's Class Limit Formula to calculate these values. This approach is also the one used by Weber and is totally non-elementary.

With the background in Ramanujan's theory of elliptic functions, modular equations and class invariants we will discuss some of the results in his paper "Modular Equations and Approximations to $\pi$" in the next post.

#### 2 comments :: Ramanujan's Class Invariants

1. Thanks so much for the reply on stackexchange. I see now that Ramanujan's equations produce the relation inquired about. I was curious as to whether they could produce relations that require that k, or the exponents, be even integers as well as odds.

As an example:
e^(pi/6) = 2^(3/8) * PI [1 + e^(-2kpi)] with k = 1 to inf.

and also:
e^(pi/4) = [2*(2^.5 + 1)]^.5 * PI [1 + e^(-4kpi)] / PI [1 + e^(-2(2k-1))] with k = 1 to inf.

This second relation of e^(pi/4) with the value [2*(2^.5 +1)]^.5 uses the series (2,6,10,14,18...) and the series (4,8,12,16,20...). Having read further through your blog, I believe that the first series would be G4, with n=4, so that the odds (1,3,5,7,9...) are all multiplied by 2. However, I am not sure how to achieve the second series which requires even integers.

I am sorry for my poor formatting. Thanks again.

Anonymous

2. @Anonymous,
I have already given the solution for the formula $$e^{\pi/6} = 2^{3/8}\prod_{k = 1}^{\infty}(1 + e^{-2k\pi})$$ on MSE(http://math.stackexchange.com/a/923319/72031). Only thing to note is that this formula is wrong. We need to replace $\pi/6$ with $\pi/12$ and here $g_{4}$ is involved and not $G_{4}$ as you think.

The second product $$e^{\pi/4} = \sqrt{2(1 + \sqrt{2})}\prod_{k = 1}^{\infty}\frac{1 + e^{-4k\pi}}{1 + e^{-2(2k - 1)\pi}}$$ requires more calculation. We can proceed as follows \begin{aligned}S &= \prod_{k = 1}^{\infty}\frac{1 + e^{-4k\pi}}{1 + e^{-2(2k - 1)\pi}}\\ &= \prod_{k = 1}^{\infty}\frac{1}{1 - e^{-4(2k - 1)\pi}}\prod_{k = 1}^{\infty}\frac{1}{1 + e^{-2(2k - 1)\pi}}\\ &= \frac{1}{2^{1/4}e^{-\pi/6}g_{16}}\cdot\frac{1}{2^{1/4}e^{-\pi/12}G_{4}}\\ &= \frac{e^{\pi/4}}{2^{3/4}g_{4}G_{4}^{2}}\end{aligned} If we calculate the value of $g_{4}, G_{4}$ then we get the desired formula.