Certain Lambert Series Identities and their Proof via Trigonometry: Part 2

In the last post we saw that using a trigonometric identity Ramanujan was able to express the functions $S_{2n - 1}(x)$ or equivalently $\Phi_{0, 2n - 1}(x)$ in terms of simpler functions $P(x), Q(x), R(x)$. Continuing our journey further we start with the equation: $$\begin{align}&\left(\frac{1}{4}\cot\frac{\theta}{2} + \frac{x\sin\theta}{1 - x} + \frac{x^{2}\sin 2\theta}{1 - x^{2}} + \frac{x^{3}\sin 3\theta}{1 - x^{3}} + \cdots\right)^{2}\notag\\ &\,\,\,\,= \left(\frac{1}{4}\cot\frac{\theta}{2}\right)^{2} + \frac{x\cos\theta}{(1 - x)^{2}} + \frac{x^{2}\cos 2\theta}{(1 - x^{2})^{2}} + \frac{x^{3}\cos 3\theta}{(1 - x^{3})^{2}} + \cdots\notag\\ &\,\,\,\,+\frac{1}{2}\left\{\frac{x(1 - \cos\theta)}{1 - x} + \frac{2x^{2}(1 - \cos 2\theta)}{1 - x^{2}} + \frac{3x^{3}(1 - \cos 3\theta)}{1 - x^{3}} + \cdots\right\}\tag{1}\end{align}$$ which was established in the last post.

Expression of $\Phi_{1, 2n}(x)$ in terms of $P, Q, R$

Using the above identity we will now express the functions $\Phi_{1, 2n}(x)$ in terms of $P, Q, R$. To do so we need to express both the LHS and RHS of $(1)$ in series of powers of $\theta$. From the last post we can see that $$\begin{align}\frac{1}{4}\cot\frac{\theta}{2} &= \frac{1}{2\theta} - \frac{B_{2}}{2\cdot 2}\cdot\frac{\theta}{1!} + \frac{B_{4}}{2\cdot 4}\cdot\frac{\theta^{3}}{3!} - \cdots\notag\\ &= \frac{1}{2\theta} + E_{1}\frac{\theta}{1!} - E_{3}\frac{\theta^{3}}{3!} + E_{5}\frac{\theta^{5}}{5!} - \cdots\notag\\ \frac{x^{m}}{1 - x^{m}}\sin m\theta &= \frac{x^{m}}{1 - x^{m}}\left(m\theta - \frac{m^{3}\theta^{3}}{3!} + \frac{m^{5}\theta^{5}}{5!} - \cdots\right)\notag\end{align}$$ and therefore the LHS of $(1)$ can be written as: $$\left(\frac{1}{2\theta} + \frac{\theta}{1!}S_{1}(x) - \frac{\theta^{3}}{3!}S_{3}(x) + \frac{\theta^{5}}{5!}S_{5}(x) - \cdots\right)^{2}$$ For the RHS of $(1)$ we can see that $$\begin{align}\left(\frac{1}{4}\cot\frac{\theta}{2}\right)^{2} &= \frac{1}{16}\cot^{2}\frac{\theta}{2}\notag\\ &= \frac{1}{4\theta^{2}} - \frac{1}{24} + \frac{1}{2}\left(-\frac{B_{4}}{2\cdot 4}\frac{\theta^{2}}{2!} + \frac{B_{6}}{2\cdot 6}\frac{\theta^{4}}{4!} - \frac{B_{8}}{2\cdot 8}\frac{\theta^{6}}{6!} + \cdots\right)\notag\\ &= \frac{1}{4\theta^{2}} + E_{1} + \frac{1}{2}\left(E_{3}\frac{\theta^{2}}{2!} - E_{5}\frac{\theta^{4}}{4!} + \cdots\right)\notag\end{align}$$ and $$\begin{align}\frac{mx^{m}(1 - \cos m\theta)}{1 - x^{m}} &= \frac{mx^{m}}{1 - x^{m}}\left(\frac{m^{2}\theta^{2}}{2!} - \frac{m^{4}\theta^{4}}{4!} + \cdots\right)\notag\\ \frac{x^{m}\cos m\theta}{(1 - x^{m})^{2}} &= \frac{x^{m}}{(1 - x^{m})^{2}}\left(1 - \frac{m^{2}\theta^{2}}{2!} + \frac{m^{4}\theta^{4}}{4!} - \cdots \right)\notag\end{align}$$ Noting that $$\begin{align}\sum_{m = 1}^{\infty}\frac{m^{r}x^{m}}{(1 - x^{m})^{2}} &= \sum_{m = 1}^{\infty}m^{r}x^{m}\sum_{n = 1}^{\infty}nx^{mn - m}\notag\\ &= \sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty}m^{r}nx^{mn} = \Phi_{1, r}(x)\notag\end{align}$$ the RHS of equation $(1)$ can be written as: $$\begin{align}\frac{1}{4\theta^{2}} &+ E_{1} + \Phi_{1, 0}(x) - \frac{\theta^{2}}{2!}\Phi_{1, 2}(x) + \frac{\theta^{4}}{4!}\Phi_{1, 4}(x) - \frac{\theta^{6}}{6!}\Phi_{1, 6}(x) + \cdots\notag\\ &+ \frac{1}{2}\left(\frac{\theta^{2}}{2!}S_{3}(x) - \frac{\theta^{4}}{4!}S_{5}(x) + \cdots\right)\notag\end{align}$$ Thus the equation $(1)$ is transformed into: $$\begin{align}&\left(\frac{1}{2\theta} + \frac{\theta}{1!}S_{1}(x) - \frac{\theta^{3}}{3!}S_{3}(x) + \frac{\theta^{5}}{5!}S_{5}(x) - \cdots\right)^{2}\notag\\ &\,\,\,\,\,\,\,\,=\frac{1}{4\theta^{2}} + S_{1}(x) - \frac{\theta^{2}}{2!}\Phi_{1, 2}(x) + \frac{\theta^{4}}{4!}\Phi_{1, 4}(x) - \frac{\theta^{6}}{6!}\Phi_{1, 6}(x) + \cdots\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+ \frac{1}{2}\left(\frac{\theta^{2}}{2!}S_{3}(x) - \frac{\theta^{4}}{4!}S_{5}(x) + \cdots\right)\notag\end{align}$$ For even positive interger $n$ if we equate the coefficients of $\theta^{n}$ on both sides of the above equation we get: $$\begin{align}\frac{n + 3}{2(n + 1)}S_{n + 1}(x) - \Phi_{1, n}(x) &= \binom{n}{1}S_{1}(x)S_{n - 1}(x)\notag\\ &+ \binom{n}{3}S_{3}(x)S_{n - 3}(x) + \cdots\notag\\ &+ \binom{n}{n - 1}S_{n - 1}(x)S_{1}(x)\tag{2}\end{align}$$ Putting $n = 2$ we get $$\begin{align}\frac{5}{6}S_{3}(x) - \Phi_{1, 2}(x) &= 2S_{1}^{2}(x)\notag\\ \Rightarrow \frac{5}{6}\frac{Q}{240} - \Phi_{1, 2}(x) &= 2\left(\frac{-P}{24}\right)^{2}\notag\end{align}$$ or $$288\Phi_{1, 2}(x) = Q - P^{2}\tag{3}$$ Similarly with $n = 4, 6$ we get $$\begin{align}720\Phi_{1, 4}(x) &= PQ - R\tag{4}\\ 1008\Phi_{1, 6}(x) &= Q^{2} - PR\tag{5}\end{align}$$ Now that we have expressed $\Phi_{1, 2n}(x)$ in terms of $P, Q, R$ we proceed to handle the case for general $\Phi_{r, s}(x)$.

Expression of $\Phi_{r, s}(x)$ in terms of $P, Q, R$

We know that $$P = 1 - 24\Phi_{1, 0}(x) = 1 - 24\sum_{n = 1}^{\infty}\sigma_{1}(n)x^{n}$$ and therefore $$x\frac{dP}{dx} = -24\sum_{n = 1}^{\infty}n\sigma_{1}(n)x^{n} = -24\Phi_{1, 2}(x)$$ and using $(3)$ we get $$x\frac{dP}{dx} = \frac{P^{2} - Q}{12}\tag{6}$$ Similarly we obtain: $$\begin{align}x\frac{dQ}{dx} &= \frac{PQ - R}{3}\tag{7}\\ x\frac{dR}{dx} &= \frac{PR - Q^{2}}{2}\tag{8}\end{align}$$ Again it is easy to verify (via repeated differentiation) that $$\Phi_{r, s}(x) = \left(x\frac{d}{dx}\right)^{r}\Phi_{0, s - r}(x)\tag{9}$$ If $s - r$ is odd (i.e. $r$ and $s$ are of different parity) then $\Phi_{0, s - r}(x)$ can be expressed as a polynomial in $P, Q, R$ and thereby using $(6), (7), (8)$ we can see from the above equation that $\Phi_{r, s}(x)$ can be expressed as a polynomial in $P, Q, R$.

We show one calculation for $\Phi_{2, 3}(x)$. Clearly from $(9)$ we have $$\begin{align} \Phi_{2, 3}(x) &= \left(x\frac{d}{dx}\right)^{2}\Phi_{0, 1}(x)\notag\\ &= x\frac{d}{dx}\Phi_{1, 2}(x)\notag\\ &= x\frac{d}{dx}\left(\frac{Q - P^{2}}{288}\right)\notag\\ &= \frac{x}{288}\left(\frac{dQ}{dx} - 2P\frac{dP}{dx}\right)\notag\\ &= \frac{1}{288}\left(\frac{PQ - R}{3} - \frac{P^{3} - PQ}{6}\right)\notag\\ &= \frac{3PQ - 2R - P^{3}}{1728}\notag\end{align}$$ Using equations $(6), (7), (8)$ we can see that $$\begin{align}x\frac{d}{dx}(Q^{3} - R^{2}) &= 3Q^{2}x\frac{dQ}{dx} - 2Rx\frac{dR}{dx}\notag\\ &= 3Q^{2}\frac{PQ - R}{3} - 2R\frac{PR - Q^{2}}{2}\notag\\ &= PQ^{3} - RQ^{2} - PR^{2} + RQ^{2}\notag\\ &= P(Q^{3} - R^{2})\notag\end{align}$$ and hence it follows that $$x\frac{d}{dx}\{\log(Q^{3} - R^{2})\} = P\tag{10}$$ Aagain if we note the definition of eta function $\eta(x)$ given by $$\eta(x) = x^{1/24}(1 - x)(1 - x^{2})(1 - x^{3})(1 - x^{4})\cdots$$ it is clear that $$P = 24x\frac{d}{dx}\{\log(\eta(x))\}$$ or $$P = x\frac{d}{dx}\{\log(\eta^{24}(x))\}\tag{11}$$ From $(10), (11)$ it follows that we must have $$Q^{3} - R^{2} = A\eta^{24}(x)$$ where $A$ is some constant. Since we have $$Q = 1 + 240x + \cdots,\,\, R = 1 - 504x + \cdots$$ it follows that $$Q^{3} - R^{2} = (3\cdot 240 + 2\cdot 504)x + \cdots = 1728x + \cdots$$ and therefore the constant $A$ must be $1728$.

We finally arrive at the beautiful identity: $$Q^{3} - R^{2} = 1728x\{(1 - x)(1 - x^{2})(1 - x^{3})\cdots\}^{24}\tag{12}$$

Relation of $P, Q, R$ with Elliptic Integrals $K, E$

Using the differential equations $(6), (7), (8)$ we can deduce many further properties of $P, Q, R$. Here we will establish their link with the elliptic integrals $K, E$ and modulus $k$. We know that $$\frac{dK}{dk} = \frac{E - k'^{2}K}{kk'^{2}},\,\,\frac{dE}{dk} = \frac{E - K}{k}$$ Eliminating $E$ from the above relations we get $$kk'^{2}\frac{d^{2}K}{dk^{2}} + (1 - 3k^{2})\frac{dK}{dk} - kK = 0\tag{13}$$ Now if we set $x = q^{2}$ we find that $$\begin{align} P(q^{2}) &= 12q\frac{d}{dq}[\log\{q^{1/12}(1 - q^{2})(1 - q^{4})(1 - q^{6})\cdots\}]\notag\\ &= 12q\frac{dk}{dq}\frac{d}{dk}\left\{\log\left(2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\right)\right\}\notag\\ &= \frac{24kk'^{2}K^{2}}{\pi^{2}}\left\{\frac{1}{2K}\frac{dK}{dk} + \frac{1}{6k} - \frac{k}{6k'^{2}}\right\}\notag\\ &= \left(\frac{2K}{\pi}\right)^{2}\left\{\frac{3kk'^{2}}{K}\cdot\frac{E - k'^{2}K}{kk'^{2}} + k'^{2} - k^{2}\right\}\notag\\ &= \left(\frac{2K}{\pi}\right)^{2}\left\{\frac{3E}{K} + k^{2} - 2\right\}\tag{14}\end{align}$$ From $(6)$ and $(14)$we see that $$\begin{align}q\frac{dP(q^{2})}{dq} &= \frac{P^{2}(q^{2}) - Q(q^{2})}{6}\notag\\ \Rightarrow Q(q^{2}) &= P^{2}(q^{2}) - 6q\frac{dP(q^{2})}{dq}\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}\left\{\frac{3E}{K} + k^{2} - 2\right\}^{2} - 6q\frac{dk}{dq}\frac{d}{dk}\{P(q^{2})\}\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}\left[\left\{\frac{3kk'^{2}}{K}\frac{dK}{dk} + 1 - 2k^{2}\right\}^{2} \right. \notag\\ &\,\,\,\,- \left. \frac{3kk'^{2}}{K^{2}}\frac{d}{dk}\left\{3kk'^{2}K\frac{dK}{dk} + K^{2}(1 - 2k^{2})\right\}\right]\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}\left[\left\{\frac{3kk'^{2}}{K}\frac{dK}{dk} + 1 - 2k^{2}\right\}^{2}\right. \notag\\ &\,\,\,\,- \frac{3kk'^{2}}{K^{2}}\left\{3kk'^{2}\left(K\frac{d^{2}K}{dk^{2}} + \left(\frac{dK}{dk}\right)^{2}\right)\right.\notag\\ &\,\,\,\,+ \left. \left. K\frac{dK}{dk}(5 - 13k^{2}) - 4kK^{2}\right\}\right]\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}\left[\left\{\frac{3kk'^{2}}{K}\frac{dK}{dk} + 1 - 2k^{2}\right\}^{2}\right. \notag\\ &\,\,\,\,- \frac{9k^{2}k'^{4}}{K}\frac{d^{2}K}{dk^{2}} - \frac{9k^{2}k'^{4}}{K^{2}}\left(\frac{dK}{dk}\right)^{2}\notag\\ &\,\,\,\,- \left. \frac{3kk'^{2}}{K}(5 - 13k^{2})\frac{dK}{dk} + 12k^{2}k'^{2}\right]\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}\left\{-\frac{9k^{2}k'^{4}}{K}\frac{d^{2}K}{dk^{2}}\right.\notag\\ &\,\,\,\,+ \left. \frac{9kk'^{2}}{K}(3k^{2} - 1)\frac{dK}{dk} + (1 - 2k^{2})^{2} + 12k^{2}k'^{2}\right\}\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}\left\{-\frac{9kk'^{2}}{K}\left(kk'^{2}\frac{d^{2}K}{dk^{2}} + (1 - 3k^{2})\frac{dK}{dk} - kK\right)\right.\notag\\ &\,\,\,\,+ \left. 1 - 4k^{2} + 4k^{4} + 3k^{2} - 3k^{4}\right\}\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}(1 - k^{2} + k^{4})\tag{15}\end{align}$$ Similarly using$(7), (14), (15)$ we get $$\begin{align} R(q^{2}) &= P(q^{2})Q(q^{2}) - \frac{3q}{2}\frac{dQ(q^{2})}{dq}\notag\\ &= \left(\frac{2K}{\pi}\right)^{6}\left\{\frac{3kk'^{2}}{K}\frac{dK}{dk} + 1 - 2k^{2}\right\}\{1 - k^{2}k'^{2}\}\notag\\ &\,\,\,\,- \frac{3kk'^{2}K^{2}}{\pi^{2}}\frac{d}{dk}\left\{\left(\frac{2K}{\pi}\right)^{4}\{1 - k^{2}k'^{2}\}\right\}\notag\\ &=\left(\frac{2K}{\pi}\right)^{6}\left[\left\{\frac{3kk'^{2}}{K}\frac{dK}{dk} + 1 - 2k^{2}\right\}\{1 - k^{2}k'^{2}\}\right.\notag\\ &\,\,\,\,- \left.\frac{3kk'^{2}}{4K^{4}}\frac{d}{dk}\left\{K^{4}\{1 - k^{2}k'^{2}\}\right\}\right]\notag\\ &= \left(\frac{2K}{\pi}\right)^{6}\left[\left\{\frac{3kk'^{2}}{K}\frac{dK}{dk} + 1 - 2k^{2}\right\}\{1 - k^{2}k'^{2}\}\right.\notag\\ &\,\,\,\,- \left.\frac{3kk'^{2}}{4K^{4}}\left\{K^{4}(4k^{3} - 2k) + 4K^{3}\frac{dK}{dk}(1 - k^{2}k'^{2})\right\}\right]\notag\\ &= \left(\frac{2K}{\pi}\right)^{6}\left[\left\{\frac{3kk'^{2}}{K}\frac{dK}{dk} + 1 - 2k^{2}\right\}\{1 - k^{2}k'^{2}\}\right.\notag\\ &\,\,\,\,+ \left.\frac{3k^{2}k'^{2}}{2}(1 - 2k^{2}) -\frac{3kk'^{2}}{K}\frac{dK}{dk}(1 - k^{2}k'^{2})\right]\notag\\ &= \left(\frac{2K}{\pi}\right)^{6}(1 - 2k^{2})\left(1 + \frac{k^{2}k'^{2}}{2}\right)\notag\\ &= \left(\frac{2K}{\pi}\right)^{6}(1 + k^{2})(1 - 2k^{2})\left(1 - \frac{k^{2}}{2}\right)\tag{16}\end{align}$$ Going further we can replace $q^{2}$ by $q$ in $(15), (16)$ and note that this leads to replacing $K$ by $(1 + k)K$ and $k$ by $2\sqrt{k}/(1 + k)$ (see Landen's Transformation of second order). When this is done we get: $$\begin{align} Q(q) &= \left(\frac{2(1 + k)K}{\pi}\right)^{4}\left(1 - \frac{4k}{(1 + k)^{2}} + \frac{16k^{2}}{(1 + k)^{4}}\right)\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}\{(1 + k)^{4} - 4k(1 + k)^{2} + 16k^{2}\}\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}\{(1 + k)^{2}(1 - k)^{2} + 16k^{2}\}\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}(1 + 14k^{2} + k^{4})\tag{17}\end{align}$$ and $$\begin{align} R(q) &= \left(\frac{2(1 + k)K}{\pi}\right)^{6}\left(1 + \frac{4k}{(1 + k)^{2}}\right)\notag\\ &\,\,\,\,\,\,\,\,\left(1 - \frac{8k}{(1 + k)^{2}}\right)\left(1 - \frac{2k}{(1 + k)^{2}}\right)\notag\\ &= \left(\frac{2K}{\pi}\right)^{6}\left\{(1 + k^{2} + 6k)(1 + k^{2} - 6k)(1 + k^{2})\right\}\notag\\ &= \left(\frac{2K}{\pi}\right)^{6}\left\{((1 + k^{2})^{2} - 36k^{2})(1 + k^{2})\right\}\notag\\ &= \left(\frac{2K}{\pi}\right)^{6}(1 + k^{2})(1 - 34k^{2} + k^{4})\tag{18}\end{align}$$ To get $P(q)$ we need to first see how $\eta(q^{2})$ transforms when $q^{2}$ is replaced by $q$. Clearly we have $$\eta(q^{2}) = q^{1/12}(1 - q^{2})(1 - q^{4})(1 - q^{6})\cdots = 2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}$$ so that $$\begin{align}\eta(q) &= q^{1/24}(1 - q)(1 - q^{2})(1 - q^{3})\cdots\notag\\ &= 2^{-1/3}\sqrt{\frac{2(1 + k)K}{\pi}}\left(\frac{2\sqrt{k}}{1 + k}\sqrt{1 - \frac{4k}{(1 + k)^{2}}}\right)^{1/6}\notag\\ &= 2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{19}\end{align}$$ and then we can see that $$\begin{align} P(q) &= 24q\frac{d}{dq}\{\log \eta(q)\}\notag\\ &= \frac{48kk'^{2}K^{2}}{\pi^{2}}\frac{d}{dk}\left\{\log\left(2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}\right)\right\}\notag\\ &= 12kk'^{2}\left(\frac{2K}{\pi}\right)^{2}\left(\frac{1}{2K}\frac{dK}{dk} + \frac{1}{12k} - \frac{k}{3k'^{2}}\right)\notag\\ &= \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6kk'^{2}}{K}\frac{dK}{dk} + 1 - 5k^{2}\right)\notag\\ &= \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6E}{K} + k^{2} - 5\right)\tag{20}\end{align}$$ Another set of formulas come by replacing $q$ with $-q$ in $(17), (18), (20)$. Note that replacing $q$ with $-q$ leads to replacing $K$ with $k'K$ and $k^{2}$ with $(-k^{2}/k'^{2})$. With this understanding we have: $$\begin{align}Q(-q) &= \left(\frac{2k'K}{\pi}\right)^{4}\left(1 - \frac{14k^{2}}{k'^{2}}+ \frac{k^{4}}{k'^{4}}\right)\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}(k'^{4} - 14k^{2}k'^{2}+ k^{4})\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}\{(1 - k^{2})^{2} - 14k^{2}(1 - k^{2})+ k^{4}\}\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}(1 - 16k^{2} + 16k^{4})\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}(1 - 16k^{2}k'^{2})\notag\\ &= \left(\frac{2K}{\pi}\right)^{4}(1 - 4G^{-24})\tag{21}\end{align}$$ and $$\begin{align} R(-q) &= \left(\frac{2k'K}{\pi}\right)^{6}\left(1 - \frac{k^{2}}{k'^{2}}\right)\left(1 + 34\frac{k^{2}}{k'^{2}} + \frac{k^{4}}{k'^{4}}\right)\notag\\ &= \left(\frac{2K}{\pi}\right)^{6}(1 - 2k^{2})(k'^{4} + 34k^{2}k'^{2} + k^{4})\notag\\ &= \left(\frac{2K}{\pi}\right)^{6}(1 - 2k^{2})(1 + 32k^{2}k'^{2})\notag\\ &= \left(\frac{2K}{\pi}\right)^{6}(1 - 2k^{2})(1 + 8G^{-24})\tag{22}\end{align}$$ To get $P(-q)$ requires more work and to that end we first rewrite the relation $(20)$ in another form: $$\begin{align}P(q) &= \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6kk'^{2}}{K}\frac{dK}{dk} + 1 - 5k^{2}\right)\notag\\ &= \left(\frac{2K}{\pi}\right)^{2}\left(\frac{12k^{2}k'^{2}}{2kK}\frac{dK}{dk} + 1 - 5k^{2}\right)\notag\\ &= \left(\frac{2K}{\pi}\right)^{2}\left(\frac{12k^{2}k'^{2}}{K}\frac{dK}{d(k^{2})} + 1 - 5k^{2}\right)\notag\end{align}$$ and then we can find $P(-q)$ as follows: $$\begin{align}P(-q) &= \left(\frac{2k'K}{\pi}\right)^{2}\left\{\frac{-12k^{2}}{k'^{2}}\left(1 + \frac{k^{2}}{k'^{2}}\right)\frac{1}{k'K}\frac{d(k'K)}{d(-k^{2}/k'^{2})} + 1 + \frac{5k^{2}}{k'^{2}}\right\}\notag\\ &= \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6kk'}{K}\frac{d(k'K)}{dk} + 1 + 4k^{2}\right)\notag\\ &= \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6kk'^{2}}{K}\frac{dK}{dk} + 1 - 2k^{2}\right)\notag\\ &= \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6(E - k'^{2}K)}{K} + 1 - 2k^{2}\right)\notag\\ &= \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6E}{K} + 4k^{2} - 5\right)\tag{23}\end{align}$$ We next define another invariant $J(q)$ by: $$J(q) = \frac{Q^{3}(-q) - R^{2}(-q)}{Q^{3}(-q)} = \frac{1728\eta^{24}(-q)}{Q^{3}(-q)}$$ Now from $(19)$ we see that $$\eta^{24}(q) = 2^{-4}\left(\frac{2K}{\pi}\right)^{12}k^{2}(1 - k^{2})^{4}$$ and hence replacing $q$ by $-q$ we get $$\begin{align}\eta^{24}(-q) &= 2^{-4}\left(\frac{2k'K}{\pi}\right)^{12}\cdot\frac{-k^{2}}{k'^{2}}\left(1 + \frac{k^{2}}{k'^{2}}\right)^{4}\notag\\ &= 2^{-4}\left(\frac{2K}{\pi}\right)^{12}(-k^{2}k'^{2})\notag\\ &= -2^{-6}\left(\frac{2K}{\pi}\right)^{12}G^{-24}\notag\end{align}$$ Using $(21)$ and definition of $J(q)$ above we arrive at $$J(q) = \frac{Q^{3}(-q) - R^{2}(-q)}{Q^{3}(-q)} = \frac{-27G^{-24}}{(1 - 4G^{-24})^{3}} = \frac{-27G^{48}}{(G^{24} - 4)^{3}}\tag{24}$$ Also using expressions for $\eta^{24}(-q), Q(-q), R(-q)$ in terms of $G$ we can establish that $Q^{3}(-q) - R^{2}(-q) = 1728\eta^{24}(-q)$ thereby furnishing another proof of $(12)$.

Identities Concerning Divisor Functions

From the last post we have the following identities: $$\begin{align}1 + 480\Phi_{0, 7}(x) &= Q^{2}\tag{25}\\ 1 - 264\Phi_{0, 9}(x) &= QR\tag{26}\\ 691 + 65520\Phi_{0, 11}(x) &= 441Q^{3} + 250R^{2}\tag{27}\\ 1 - 24\Phi_{0, 13}(x) &= Q^{2}R\tag{28}\end{align}$$ First of these identities can be written as: $$1 + 480\sum_{n = 1}^{\infty}\sigma_{7}(n)x^{n} = \left(1 + 240\sum_{n = 1}^{\infty}\sigma_{3}(n)x^{n}\right)^{2}$$ Equating coefficients of $x^{n}$ on both sides we get: $$\sigma_{7}(n) = \sigma_{3}(n) + 120\sum_{i = 1}^{n - 1}\sigma_{3}(i)\sigma_{3}(n - i)$$ so that $\sigma_{7}(n) - \sigma_{3}(n)$ is divisible by $120$ for all positive integers $n$.

Similarly from $(26)$ we get: $$11\sigma_{9}(n) = \{21\sigma_{5}(n) - 10\sigma_{3}(n)\} + 5040\sum_{i = 1}^{n - 1}\sigma_{3}(i)\sigma_{5}(n - i)$$ and thus $11\sigma_{9}(n) + 10\sigma_{3}(n) - 21\sigma_{5}(n)$ is divisible by $5040$.

From $(25)$ and $(28)$ we get $$1 - 24\sum_{n = 1}^{\infty}\sigma_{13}(n)x^{n} = \left(1 + 480\sum_{n = 1}^{\infty}\sigma_{7}(n)x^{n}\right)\left(1 - 504\sum_{n = 1}^{\infty}\sigma_{5}(n)x^{n}\right)$$ and thus on equating coefficients of $x^{n}$ we get $$\sigma_{13}(n) = \{21\sigma_{5}(n) - 20\sigma_{7}(n)\} + 10080\sum_{i = 1}^{n - 1}\sigma_{5}(i)\sigma_{7}(n - i)$$ We can rewrite the equation $(27)$ as $$691 + 65520\Phi_{0, 11}(x) = 441(Q^{3} - R^{2}) + 691R^{2}$$ so that $$691 + 65520\Phi_{0, 11}(x) = 441\cdot 1728\eta^{24}(x) + 691R^{2}\tag{29}$$ Let us define Ramanujan's Tau function $\tau(n)$ by $$\sum_{n = 1}^{\infty}\tau(n)x^{n} = \eta^{24}(x) = x\prod_{n = 1}^{\infty}(1 - x^{n})^{24}$$ and then equating coefficients of $x^{n}$ in $(29)$ we get $$\begin{align}65520\sigma_{11}(n) &= 441\cdot 1728\tau(n) - 691\cdot 1008\sigma_{5}(n)\notag\\ &\,\,\,\,+ 691\cdot 504^{2}\sum_{i = 1}^{n - 1}\sigma_{5}(i)\sigma_{5}(n - i)\notag\\ \Rightarrow 65\sigma_{11}(n) &= 756\tau(n) - 691\sigma_{5}(n) + 691\cdot 252\sum_{i = 1}^{n - 1}\sigma_{5}(i)\sigma_{5}(n - i)\notag\\ \Rightarrow 65(\sigma_{11}(n) - \tau(n)) &= 691\tau(n) - 691\sigma_{5}(n) + 691\cdot 252\sum_{i = 1}^{n - 1}\sigma_{5}(i)\sigma_{5}(n - i)\notag\end{align}$$ and thus we arrive at the famous congruence satisfied by $\tau(n)$ $$\tau(n) \equiv \sigma_{11}(n) \pmod{691}\tag{30}$$
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