Plouffe's formula for $\pi$

After a little more thinking I got the function $a(q)$ in the form I wanted. We start from the equation $$\eta(q) = q^{1/24}\prod_{n = 1}^{\infty}(1 - q^{n}) = 2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}$$ Taking logs we get $$\frac{\log q}{24} + \sum_{n = 1}^{\infty}\log(1 - q^{n}) = \frac{\log k}{12} + \frac{\log k'}{3} + \frac{\log 2}{3} + \frac{1}{2}\log\left(\frac{K}{\pi}\right)$$ or $$-\frac{\pi}{24}\cdot\frac{K'}{K} - \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{n})} = \frac{\log k}{12} + \frac{\log k'}{3} + \frac{\log 2}{3} + \frac{1}{2}\log\left(\frac{K}{\pi}\right)$$ so that the function $a(q)$ is given by $$a(q) = \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{n})} = -\frac{\log k}{12} - \frac{\log k'}{3} - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{24K}\tag{1}$$ Now if we change $q$ to $q^{2}$ we need to replace $k$ by $(1 - k')/(1 + k')$ and $K$ by $K(1 + k')/2$ and the ratio $K'/K$ just doubles up so we get \begin{align}a(q^{2}) &= -\frac{1}{12}\log\left(\frac{1 - k'}{1 + k'}\right) - \frac{1}{6}\log\left(\frac{4k'}{(1 + k')^{2}}\right) - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K(1 + k')}{2\pi}\right) - \frac{\pi K'}{12K}\notag\\ &= -\frac{1}{6}\log\left(\frac{k}{1 + k'}\right) - \frac{1}{6}\log\left(\frac{4k'}{(1 + k')^{2}}\right) - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K(1 + k')}{2\pi}\right) - \frac{\pi K'}{12K}\notag\\ &= -\frac{1}{6}\log\left(\frac{4kk'}{(1 + k')^{3}}\right) - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K(1 + k')}{2\pi}\right) - \frac{\pi K'}{12K}\notag\\ &= -\frac{\log(4kk')}{6} - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K}{2\pi}\right) - \frac{\pi K'}{12K}\notag\\ &= -\frac{\log(kk')}{6} - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{12K}\notag \end{align} To summarize $$a(q^{2}) = -\frac{\log(kk')}{6} - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{12K}\tag{2}$$ Again replacing $q$ by $q^{2}$ we get \begin{align} a(q^{4}) &= -\frac{1}{6}\log\left(\frac{1 - k'}{1 + k'}\right) - \frac{1}{12}\log\left(\frac{4k'}{(1 + k')^{2}}\right) - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K(1 + k')}{2\pi}\right) - \frac{\pi K'}{6K}\notag\\ &= -\frac{1}{3}\log\left(\frac{k}{1 + k'}\right) - \frac{1}{12}\log\left(\frac{4k'}{(1 + k')^{2}}\right) - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K(1 + k')}{2\pi}\right) - \frac{\pi K'}{6K}\notag\\ &= -\frac{1}{12}\log(4k^{4}k') - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{2\pi}\right) - \frac{\pi K'}{6K}\notag\\ &= -\frac{1}{12}\log(k^{4}k') + \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{6K}\notag \end{align} so that $$a(q^{4}) = -\frac{1}{12}\log(k^{4}k') + \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{6K}\tag{3}$$ Now time to do some algebra. Let us put $q = e^{-\pi}$ so that $k = k' = 2^{-1/2}$ and $K' = K$. We then get $$a(q) = -\frac{\log 2}{8} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{24}$$ and $$a(q^{2}) = -\frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{12}$$ and $$a(q^{4}) = \frac{3\log 2}{8} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{6}$$ so we get $$72a(q) - 96a(q^{2}) + 24a(q^{4}) = -3\pi + 8\pi - 4\pi = \pi$$ Notice that the other terms cancel very nicely.

Note: The above derivation requires some knowledge of elliptic integrals and their connection with the theta functions. The technique of going from $q$ to $q^{2}$ is the famous Landen Transformation. Material related to these nice theories is presented in my blog and the interested reader may search elliptic integral, theta functions and Landen in the archives page.