Certain Lambert Series Identities and their Proof via Trigonometry: Part 1

Introduction

This is yet another post based on a paper of Ramanujan titled "On certain arithmetical functions" which appeared in Transactions of the Cambridge Philosophical Society in 1916. In this paper Ramanujan provided a lot of identities concerning Lambert series and thereby deduced many relations between various divisor functions. Apart from the amazing results proved in this paper, what I liked most is the very elementary approach followed by Ramanujan compared to the methods of modern authors who are seduced by the modular form.

Ramanujan's Functions $P, Q, R$

Ramanujan introduced the following Lambert series and used them extensively in deriving many identities in elliptic function theory: \begin{align}P(q) &= 1 - 24\left(\frac{q^{2}}{1 - q^{2}} + \frac{2q^{4}}{1 - q^{4}} + \frac{3q^{6}}{1 - q^{6}} + \cdots\right)\notag\\ &= 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}}\tag{1}\\ Q(q) &= 1 + 240\left(\frac{q^{2}}{1 - q^{2}} + \frac{2^{3}q^{4}}{1 - q^{4}} + \frac{3^{3}q^{6}}{1 - q^{6}} + \cdots\right)\notag\\ &= 1 + 240\sum_{n = 1}^{\infty}\frac{n^{3}q^{2n}}{1 - q^{2n}}\tag{2}\\ R(q) &= 1 - 504\left(\frac{q^{2}}{1 - q^{2}} + \frac{2^{5}q^{4}}{1 - q^{4}} + \frac{3^{5}q^{6}}{1 - q^{6}} + \cdots\right)\notag\\ &= 1 - 504\sum_{n = 1}^{\infty}\frac{n^{5}q^{2n}}{1 - q^{2n}}\tag{3}\end{align} We have already met $P(q)$ in a previous post regarding series for $1/\pi$. Also the $R(q)$ is not to be confused with the Rogers-Ramanujan continued fractions introduced in the last post. Ramanujan also used the alternative notation $L, M, N$ instead of $P, Q, R$. Again to simplify matters regarding manipulation of above series Ramanujan used the variable $x = q^{2}$ and hence we get the following notation which will be used subsequently in this post: \begin{align}P(x) &= 1 - 24\left(\frac{x}{1 - x} + \frac{2x^{2}}{1 - x^{2}} + \frac{3x^{3}}{1 - x^{3}} + \cdots\right)\notag\\ &= 1 - 24\sum_{n = 1}^{\infty}\frac{nx^{n}}{1 - x^{n}}\tag{4}\\ Q(x) &= 1 + 240\left(\frac{x}{1 - x} + \frac{2^{3}x^{2}}{1 - x^{2}} + \frac{3^{3}x^{3}}{1 - x^{3}} + \cdots\right)\notag\\ &= 1 + 240\sum_{n = 1}^{\infty}\frac{n^{3}x^{n}}{1 - x^{n}}\tag{5}\\ R(x) &= 1 - 504\left(\frac{x}{1 - x} + \frac{2^{5}x^{2}}{1 - x^{2}} + \frac{3^{5}x^{3}}{1 - x^{3}} + \cdots\right)\notag\\ &= 1 - 504\sum_{n = 1}^{\infty}\frac{n^{5}x^{n}}{1 - x^{n}}\tag{6}\end{align} The Lambert series above can be easily expressed as generating functions for divisor function. In general for any positive integer $r$, we can see that $$\sum_{n = 1}^{\infty}\frac{n^{r}x^{n}}{1 - x^{n}} = \sum_{n = 1}^{\infty}\sigma_{r}(n)x^{n}$$ where $\sigma_{r}(n)$ denotes sum of $r^{\text{th}}$ powers of divisors of $n$.

Ramanujan considered these sums and its generalization below: \begin{align}S_{r}(x) &= E_{r} + \frac{1^{r}x}{1 - x} + \frac{2^{r}x^{2}}{1 - x^{2}} + \frac{3^{r}x^{3}}{1 - x^{3}} + \cdots\notag\\ &= E_{r} + \sum_{n = 1}^{\infty}\frac{n^{r}x^{n}}{1 - x^{n}}\notag\\ &= E_{r} + \sum_{n = 1}^{\infty}\sigma_{r}(n)x^{n}\tag{7}\\ \Phi_{r, s}(x) &= \sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty}m^{r}n^{s}x^{mn}\tag{8}\end{align} where $E_{r}$ is a suitably chosen constant related with Bernoulli's numbers as we shall see later. It is easy to see that \begin{align} \Phi_{r, s}(x) &= \Phi_{s, r}(x) = \sum_{n = 1}^{\infty}n^{s}\sigma_{r - s}(n)x^{n}\tag{9}\\ S_{r}(x) &= E_{r} + \Phi_{0, r}(x)\tag{10}\end{align} Ramanujan was able to express $\Phi_{r, s}(x)$, with $r + s$ an odd positive integer, in terms of $P, Q, R$ in a very elementary manner using trigonometrical series. This is one of the truly amazing proofs which Ramanujan provided. In a way the proof shows that a lot more can be achieved with elementary stuff than people think. To understand the proof we need to develop the series for $\cot(x)$ and there we will see the use of Bernoulli's numbers.

Expansion of $\cot x$

We know that the Bernoulli's numbers $B_{n}$ are defined by $$\frac{x}{e^{x} - 1} = \sum_{n = 0}^{\infty}B_{n}\frac{x^{n}}{n!}\tag{11}$$ Now we can see that $$\frac{x}{e^{x} - 1} = \dfrac{1}{1 + \dfrac{x}{2!} + \dfrac{x^{2}}{3!} + \cdots} = 1 - \frac{x}{2} + \cdots$$ and it is easy to verify that $\{x/(e^{x} - 1)\} + (x/2)$ is an even function therefore it follows that: $$B_{0} = 1, B_{1} = \frac{-1}{2}, B_{2} = \frac{1}{6}, B_{4} = \frac{-1}{30}, B_{3} = B_{5} = \cdots = B_{2n + 1} = 0\tag{12}$$ Next we proceed to find expansion of $\cot x$ in powers of $x$. We have \begin{align} \cot x &= \frac{\cos x}{\sin x} = i\cdot \frac{e^{ix} + e^{-ix}}{e^{ix} - e^{-ix}} = i\cdot\frac{e^{2ix} + 1}{e^{2ix} - 1} = i\cdot\left(1 + \frac{2}{e^{2ix} - 1}\right)\notag\\ &= i + \frac{1}{x}\cdot\frac{2ix}{e^{2ix} - 1}\notag\\ &= i + \frac{1}{x}\left(1 - \frac{2ix}{2} + B_{2}\cdot\frac{(2ix)^{2}}{2!} + B_{4}\cdot\frac{(2ix)^{4}}{4!} + \cdots\right)\notag\\ &= \frac{1}{x} - B_{2}\frac{2^{2}x}{2!} + B_{4}\frac{2^{4}x^{3}}{4!} - B_{6}\frac{2^{6}x^{5}}{6!} + \cdots\tag{13}\end{align} Differentiating with respect to $x$ we get \begin{align} -\operatorname{cosec}^{2} x &= -\frac{1}{x^{2}} - B_{2}\frac{2^{2}}{2!} + 3B_{4}\frac{2^{4}x^{2}}{4!} - 5B_{6}\frac{2^{6}x^{4}}{6!} + \cdots\notag\\ \Rightarrow 1 + \cot^{2} x &= \frac{1}{x^{2}} + \frac{1}{3} - 3B_{4}\frac{2^{4}x^{2}}{4!} + \cdots\notag\\ \Rightarrow \cot^{2} x &= \frac{1}{x^{2}} - \frac{2}{3} - 3B_{4}\frac{2^{4}x^{2}}{4!} + 5B_{6}\frac{2^{6}x^{4}}{6!} - 7B_{8}\frac{2^{8}x^{6}}{8!} + \cdots\tag{14}\end{align}

A Trigonometrical Identity

Ramanujan next uses the formula for sum of cosines of angles in arithmetic progression in the following manner \begin{align}2\{\cos\theta + \cos 2\theta + \cdots + \cos(n - 1)\theta\} &= \dfrac{2\cos\dfrac{n\theta}{2}\sin\dfrac{(n - 1)\theta}{2}}{\sin\dfrac{\theta}{2}}\notag\\ &= \dfrac{\sin\left(n - \dfrac{1}{2}\right)\theta - \sin \dfrac{\theta}{2}}{\sin\dfrac{\theta}{2}}\notag\\ &= \dfrac{\sin n\theta \cos\dfrac{\theta}{2} - \cos n\theta\sin\dfrac{\theta}{2}}{\sin\dfrac{\theta}{2}} - 1\notag\\ &= \cot\frac{\theta}{2}\sin n\theta - \cos n\theta - 1\notag\end{align} to get $$\cot\frac{\theta}{2}\sin n\theta = 1 + 2\cos\theta + 2\cos 2\theta + \cdots + 2\cos(n - 1)\theta + \cos n\theta\tag{15}$$ Out of the blue Ramanujan now sets out to consider the expression $$S = \left(\frac{1}{4}\cot\frac{\theta}{2} + \frac{x\sin\theta}{1 - x} + \frac{x^{2}\sin 2\theta}{1 - x^{2}} + \frac{x^{3}\sin 3\theta}{1 - x^{3}} + \cdots\right)^{2}$$ Let $u_{m} = x^{m}/(1 - x^{m})$ and then the above expression can be written as \begin{align}S &= \left(\frac{1}{4}\cot\frac{\theta}{2} + u_{1}\sin\theta + u_{2}\sin 2\theta + u_{3}\sin 3\theta + \cdots\right)^{2}\notag\\ &= \left(\frac{1}{4}\cot\frac{\theta}{2} + \sum_{m = 1}^{\infty}u_{m}\sin m\theta\right)^{2}\notag\\ &= \left(\frac{1}{4}\cot\frac{\theta}{2}\right)^{2} + \frac{1}{2}\cot\frac{\theta}{2}\sum_{m = 1}^{\infty}u_{m}\sin m\theta + \left(\sum_{m = 1}^{\infty}u_{m}\sin m\theta\right)^{2}\notag\\ &= \left(\frac{1}{4}\cot\frac{\theta}{2}\right)^{2} + T_{1} + T_{2}\notag\end{align} where using $(14)$ we can express $T_{1}$ in terms of cosines as: $$T_{1} = \sum_{m = 1}^{\infty}u_{m}\left\{\frac{1}{2} + \cos\theta + \cos 2\theta + \cdots + \cos (m - 1)\theta + \frac{1}{2}\cos m\theta\right\}$$ and $T_{2}$ can also be expressed in terms of cosines as follows: \begin{align}T_{2} &= \sum_{m = 1}^{\infty}u_{m}\sin m\theta\sum_{n = 1}^{\infty}u_{n}\sin n\theta\notag\\ &= \frac{1}{2}\sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty}\{\cos(m - n)\theta - \cos(m + n)\theta\}u_{m}u_{n}\notag\end{align} and thus $T_{1} + T_{2}$ can be arranged in a series of the form $$T_{1} + T_{2} = \sum_{k = 0}^{\infty}C_{k}\cos k\theta$$ Now it is clear that the contribution to $C_{0}$ from $T_{1}$ is $(1/2)\sum u_{m}$ and from $T_{2}$ the contribution is $(1/2)\sum u_{m}^{2}$ and therefore \begin{align}C_{0} &= \frac{1}{2}\sum_{m = 1}^{\infty}u_{m} + \frac{1}{2}\sum_{m = 1}^{\infty}u_{m}^{2}\notag\\ &= \frac{1}{2}\sum_{m = 1}^{\infty}u_{m}(1 + u_{m})\notag\\ &= \frac{1}{2}\sum_{m = 1}^{\infty}\frac{x^{m}}{(1 - x^{m})^{2}}\notag\\ &= \frac{1}{2}\sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty}nx^{mn}\notag\\ &= \frac{1}{2}\sum_{n = 1}^{\infty}\frac{nx^{n}}{1 - x^{n}} = \frac{1}{2}\sum_{n = 1}^{\infty}nu_{n}\notag\end{align} For $k > 0$, the part of $C_{k}$ coming from $T_{1}$ is $$\frac{1}{2}u_{k} + \sum_{m = k + 1}^{\infty}u_{m} = \frac{1}{2}u_{k} + \sum_{l = 1}^{\infty}u_{k + l}$$ and the part of $C_{k}$ coming from $T_{2}$ is $$\frac{1}{2}\sum_{m - n = k}u_{m}u_{n} + \frac{1}{2}\sum_{n - m = k}u_{m}u_{n} - \frac{1}{2}\sum_{m + n = k}u_{m}u_{n} = \sum_{l = 1}^{\infty}u_{l}u_{k + l} - \frac{1}{2}\sum_{l = 1}^{k - 1}u_{l}u_{k - l}$$ Thus it follows that for $k > 0$ \begin{align} C_{k} &= \frac{1}{2}u_{k} + \sum_{l = 1}^{\infty}u_{k + l} + \sum_{l = 1}^{\infty}u_{l}u_{k + l} - \frac{1}{2}\sum_{l = 1}^{k - 1}u_{l}u_{k - l}\notag\\ &= \frac{1}{2}u_{k} + \sum_{l = 1}^{\infty}u_{k + l}(1 + u_{l}) - \frac{1}{2}\sum_{l = 1}^{k - 1}u_{l}u_{k - l}\notag\\ &= \frac{1}{2}u_{k} + \sum_{l = 1}^{\infty}u_{k}(u_{l} - u_{k + l}) - \frac{1}{2}\sum_{l = 1}^{k - 1}u_{k}(1 + u_{l} + u_{k - 1})\notag\\ &= u_{k}\left\{\frac{1}{2} + \sum_{l = 1}^{\infty}(u_{l} - u_{k + l}) - \frac{1}{2}\sum_{l = 1}^{k - 1}(1 + u_{l} + u_{k - l})\right\}\notag\\ &= u_{k}\left\{\frac{1}{2} + u_{1} + u_{2} + \cdots + u_{k} - \frac{k - 1}{2} - (u_{1} + u_{2} + \cdots + u_{k - 1})\right\}\notag\\ &= u_{k}\left\{1 + u_{k} - \frac{k}{2}\right\}\notag\end{align} The crucial part in the above proof are the easily verifiable identities: $$u_{k + l}(1 + u_{l}) = u_{k}(u_{l} - u_{k + l}),\,\,\, u_{l}u_{k - l} = u_{k}(1 + u_{l} + u_{k - l})$$ Finally putting all the pieces together we can see that \begin{align}S &= \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{1}{2}\sum_{n = 1}^{\infty}nu_{n} + \sum_{k = 1}^{\infty}u_{k}\left(1 + u_{k} - \frac{k}{2}\right)\cos k\theta\notag\\ &= \left(\frac{1}{4}\cot\frac{\theta}{2}\right)^{2} + \sum_{m = 1}^{\infty}u_{m}(1 + u_{m})\cos m\theta + \frac{1}{2}\sum_{m = 1}^{\infty}mu_{m}(1 - \cos m\theta)\notag\\ &= \left(\frac{1}{4}\cot\frac{\theta}{2}\right)^{2} + \sum_{m = 1}^{\infty}\frac{x^{m}\cos m\theta}{(1 - x^{m})^{2}} + \frac{1}{2}\sum_{m = 1}^{\infty}\frac{mx^{m}}{1 - x^{m}}(1 - \cos m\theta)\notag\end{align} Ramanujan proved the above without using the symbols $\sum$ and $u_{m}$ and expressed his formula directly as: \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{16}\end{align} The presentation we have offered above is from G. H. Hardy's Ramanujan: Twelve Lectures on Subjects Suggested by his Life and Work. The proof above may look complicated because of symbolism, but in reality it involves basic algebraic manipulations.

In a similar manner Ramanujan uses the trigonometric identity: \begin{align}&\cot^{2}\frac{\theta}{2}(1 - \cos n\theta)\notag\\ &\,\,\,\,\,\,\,\,= (2n - 1) + 4(n - 1)\cos\theta + 4(n - 2)\cos 2\theta + \cdots\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+ 4\cos(n - 1)\theta + \cos n\theta\tag{17}\end{align} and establishes the following result: \begin{align}&\left\{\frac{1}{8}\cot^{2}\frac{\theta}{2} + \frac{1}{12} + \frac{x(1 - \cos\theta)}{1 - x}\right.\notag\\ &+ \left.\frac{2x^{2}(1 - \cos 2\theta)}{1 - x^{2}} + \frac{3x^{3}(1 - \cos 3\theta)}{1 - x^{3}} + \cdots\right\}^{2}\notag\\ &\,\,\,\,\,\,\,\,= \left(\frac{1}{8}\cot^{2}\frac{\theta}{2} + \frac{1}{12}\right)^{2}\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+ \frac{1}{12}\left\{\frac{1^{3}x(5 + \cos\theta)}{1 - x} + \frac{2^{3}x^{2}(5 + \cos 2\theta)}{1 - x^{2}}\right.\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. + \frac{3^{3}x^{3}(5 + \cos 3\theta)}{1 - x^{3}} + \cdots\right\}\tag{18}\end{align} The algebraic manipulations in this case are of similar nature but bit more complicated and hence will not be presented here. From $(14)$ it is clear that $$\frac{1}{8}\cot^{2}\frac{\theta}{2} + \frac{1}{12} = \frac{1}{2}\left(\frac{1}{\theta^{2}} - \frac{3B_{4}}{4!}\theta^{2} + \frac{5B_{6}}{6!}\theta^{4} - \frac{7B_{8}}{8!}\theta^{6} + \cdots\right)\tag{19}$$ and $$\frac{mx^{m}}{1 - x^{m}}(1 - \cos m\theta) = \frac{mx^{m}}{1 - x^{m}}\left(\frac{m^{2}\theta^{2}}{2!} - \frac{m^{4}\theta^{4}}{4!} + \cdots\right)$$ Hence the LHS of $(18)$ can be written as $$\left\{\frac{1}{2\theta^{2}} + \frac{\theta^{2}}{2!}\left(-\frac{B_{4}}{2\cdot 4} + \sum_{m = 1}^{\infty}\frac{m^{3}x^{m}}{1 - x^{m}}\right) - \frac{\theta^{4}}{4!}\left(-\frac{B_{6}}{2\cdot 6} + \sum_{m = 1}^{\infty}\frac{m^{5}x^{m}}{1 - x^{m}}\right) + \cdots\right\}^{2}$$ It is now time to relate $E_{r}$ of $(7)$ with Bernoulli's numbers as follows: $$E_{r} = -\frac{B_{r + 1}}{2(r + 1)}$$ and then using $(7)$ we can write the LHS of $(18)$ as: $$\left\{\frac{1}{2\theta^{2}} + \frac{\theta^{2}}{2!}S_{3}(x) - \frac{\theta^{4}}{4!}S_{5}(x) + \cdots\right\}^{2}$$ To compute the RHS of $(18)$ we need to first square the equation $(19)$. Instead of squaring the series expansion on right of $(19)$ I prefer to use derivatives of the expansion given in $(14)$. Thus on differentiating $(14)$ we get \begin{align}2\cot x(-1 - \cot^{2} x) &= -\frac{2}{x^{3}} - 2\cdot 3B_{4}\frac{2^{4}x}{4!} + 4\cdot 5B_{6}\frac{2^{6}x^{3}}{6!} - \cdots\notag\\ \Rightarrow \cot x + \cot^{3}x &= \frac{1}{x^{3}} + 2\cdot 3B_{4}\frac{2^{3}x}{4!} - 4\cdot 5B_{6}\frac{2^{5}x^{3}}{6!} + \cdots\notag\end{align} and using $(13)$ we get $$\cot^{3}x = \frac{1}{x^{3}} - \frac{1}{x} + \frac{2^{2}}{4!}(2\cdot 2\cdot 3B_{4} + 3\cdot 4B_{2})x - \frac{2^{4}}{6!}(2\cdot 4\cdot 5B_{6} + 5\cdot 6B_{4})x^{3} + \cdots$$ Differentiating the above equation we get \begin{align} 3\cot^{2}x(-1 - \cot^{2}x) &= -\frac{3}{x^{4}} + \frac{1}{x^{2}} + \frac{2^{2}}{4!}(2\cdot 2\cdot 3B_{4} + 3\cdot 4B_{2})\notag\\ &\,\,\,\,\,\,\,\,- \frac{2^{4}}{6!}\cdot 3(2\cdot 4\cdot 5B_{6} + 5\cdot 6B_{4})x^{2} + \cdots\notag\\ \Rightarrow 3\cot^{2}x + 3\cot^{4}x &= \frac{3}{x^{4}} - \frac{1}{x^{2}} - \frac{2^{2}}{4!}(2\cdot 2\cdot 3B_{4} + 3\cdot 4B_{2})\notag\\ &\,\,\,\,\,\,\,\,+ \frac{2^{4}}{6!}\cdot 3(2\cdot 4\cdot 5B_{6} + 5\cdot 6B_{4})x^{2} - \cdots\notag\end{align} Adding this equation and $(14)$ (and values of $B_{2}, B_{4}$) we get \begin{align} 3\cot^{4}x + 4\cot^{2}x &= \frac{3}{x^{4}} - \frac{14}{15} + \frac{2^{5}B_{6}}{6}\cdot\frac{x^{2}}{2!}\notag\\ &\,\,\,\,\,\,\,\,- \frac{2^{7}B_{8}}{8}\cdot\frac{x^{4}}{4!}\notag\\ &\,\,\,\,\,\,\,\,+ \frac{2^{9}B_{10}}{10}\cdot\frac{x^{6}}{6!}- \cdots\tag{20}\end{align} Again going back to the equation $(19)$ we see that square of its LHS is given by \begin{align} \left(\frac{1}{8}\cot^{2}\frac{\theta}{2} + \frac{1}{12}\right)^{2} &= \frac{1}{3\cdot 2^{6}}\left(3\cot^{4}\frac{\theta}{2} + 4\cot^{2}\frac{\theta}{2}\right) + \frac{1}{144}\notag\\ &= \frac{1}{4\theta^{4}} + \frac{1}{480} + \frac{1}{24}\left(\frac{B_{6}}{6}\cdot \frac{\theta^{2}}{2!} - \frac{B_{8}}{8}\cdot \frac{\theta^{4}}{4!} + \cdots\right)\notag\\ &= \frac{1}{4\theta^{4}} - \frac{B_{4}}{16} + \frac{1}{24}\left(\frac{B_{6}}{6}\cdot \frac{\theta^{2}}{2!} - \frac{B_{8}}{8}\cdot \frac{\theta^{4}}{4!} + \cdots\right)\notag\\ &= \frac{1}{4\theta^{4}} + \frac{E_{3}}{2} - \frac{1}{12}\left(E_{5}\cdot \frac{\theta^{2}}{2!} - E_{7}\cdot \frac{\theta^{4}}{4!} + \cdots\right)\notag\end{align} We can now clearly see that the RHS of $(18)$ is given by $$\frac{1}{4\theta^{4}} + \frac{S_{3}(x)}{2} - \frac{1}{12}\left(\frac{\theta^{2}}{2!}S_{5}(x) - \frac{\theta^{4}}{4!}S_{7}(x) + \cdots\right)$$ and finally the equation $(18)$ is transformed into \begin{align}&\left\{\frac{1}{2\theta^{2}} + \frac{\theta^{2}}{2!}S_{3}(x) - \frac{\theta^{4}}{4!}S_{5}(x) + \cdots\right\}^{2}\notag\\ &\,\,\,\,\,\,\,\,= \frac{1}{4\theta^{4}} + \frac{S_{3}(x)}{2} - \frac{1}{12}\left(\frac{\theta^{2}}{2!}S_{5}(x) - \frac{\theta^{4}}{4!}S_{7}(x) + \cdots\right)\tag{21}\end{align} For even integer $n > 2$ we equate the coefficients of $\theta^{n}$ on both sides to obtain the following euqation \begin{align}\frac{(n - 2)(n + 5)}{12(n + 1)(n + 2)}S_{n + 3}(x) &= \binom{n}{2} S_{3}(x)S_{n - 1}(x)\notag\\ &\,\,\,\,\,\,\,\,+ \binom{n}{4} S_{5}(x)S_{n - 3}(x) + \cdots\notag\\ &\,\,\,\,\,\,\,\,+ \binom{n}{n - 2}S_{n - 1}(x)S_{3}(x)\tag{22}\end{align} where $$\binom{n}{r} = \frac{n!}{r!(n - r)!}$$ is the usual binomial coefficient.

Now it is easy to see that we have $$P(x) = -24S_{1}(x), Q(x) = 240S_{3}(x), R(x) = -504S_{5}(x)$$ and hence using the relation $(22)$ we can evaluate $S_{2n - 1}(x)$ in terms of $P, Q, R$ for all integers $n > 0$. For small values of $n$ it is easy to apply the formula and derive the following: \begin{align} 1 + 480\Phi_{0, 7}(x) &= Q^{2}\notag\\ 1 - 264\Phi_{0, 9}(x) &= QR\notag\\ 691 + 65520\Phi_{0, 11}(x) &= 441Q^{3} + 250R^{2}\notag\\ 1 - 24\Phi_{0, 13}(x) &= Q^{2}R\notag\end{align} Ramanujan does not stop here and actually uses the relation $(22)$ to evaluate $S_{7}(x), S_{9}(x), \ldots, S_{31}(x)$ and gives his results in terms of the function $\Phi_{r, s}(x)$. Ramanujan did all this for his love of numbers and we show one example here which sheds light on the nature of numbers he dealt with: \begin{align} 7709321041217+32640\Phi_{0, 31}(x) &= 764412173217Q^{8}(x)\notag\\ &\,\,\,\,\,\,\,\,+ 5323905468000Q^{5}(x)R^{2}(x)\notag\\ &\,\,\,\,\,\,\,\,+ 1621003400000Q^{2}(x)R^{4}(x)\notag \end{align} In the next post we will analyze the equation $(16)$ and the results derived from it.

Postscript: L. C. Shen provided another proof of $(16)$ in 1993 using derivatives of theta functions which we reproduce below:

We have from these posts \begin{align} \theta_{1}(z, q) &= 2q^{1/4}\sin z \prod_{n = 1}^{\infty}(1 - q^{2n})(1 - 2q^{2n}\cos 2z + q^{4n})\notag\\ &= 2q^{1/4}\sum_{n = 0}^{\infty}(-1)^{n}q^{n(n + 1)}\sin(2n + 1)z\tag{23}\end{align} Differentiating twice the infinite series representation of $\theta_{1}(z, q)$ with respect to $z$ we get $$\frac{\partial^{2}\theta_{1}}{\partial z^{2}} = -2q^{1/4}\sum_{n = 0}^{\infty}(-1)^{n}q^{n(n + 1)}(2n + 1)^{2}\sin(2n + 1)z$$ Again differentiating $\theta_{1}(z, q)$ with respect to $q$ we get \begin{align}\frac{\partial \theta_{1}}{\partial q} &= 2\sum_{n = 0}^{\infty}(-1)^{n}(n + 1/2)^{2}q^{(n + 1/2)^{2} - 1}\sin(2n + 1)z\notag\\ &= \frac{1}{2}q^{-3/4}\sum_{n = 0}^{\infty}(-1)^{n}q^{n(n + 1)}(2n + 1)^{2}\sin(2n + 1)z\notag\end{align} and thus we arrive at the partial differential equation satisfied by $\theta_{1}$ $$\frac{\partial^{2}\theta_{1}}{\partial z^{2}} = -4q\frac{\partial \theta_{1}}{\partial q}\tag{24}$$ Next by logarithmic differentiation of the product expansion of $\theta_{1}(z, q)$ with respect to $z$ we get \begin{align}\frac{1}{\theta_{1}}\frac{\partial \theta_{1}}{\partial z} &= \cot z + 4\sum_{n = 1}^{\infty}\frac{q^{2n}\sin 2z}{1 - 2q^{2n}\cos 2z + q^{4n}}\notag\\ &= \cot z + \frac{4}{2i}\sum_{n = 1}^{\infty}\frac{q^{2n}(e^{2iz} - e^{-2iz})}{(1 - q^{2n}e^{2iz})(1 - q^{2n}e^{-2iz})}\notag\\ &= \cot z + \frac{4}{2i}\sum_{n = 1}^{\infty}q^{2n}\left(\frac{e^{2iz}}{1 - q^{2n}e^{2iz}} - \frac{e^{-2iz}}{1 - q^{2n}e^{-2iz}}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{n = 1}^{\infty}\frac{q^{2n}e^{2iz}}{1 - q^{2n}e^{2iz}} - \sum_{n = 1}^{\infty}\frac{q^{2n}e^{-2iz}}{1 - q^{2n}e^{-2iz}}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\{q^{2n}e^{2iz}\}^{m} - \sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\{q^{2n}e^{-2iz}\}^{m}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}q^{2m}e^{2imz}q^{2m(n - 1)}\right. \notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left. - \sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}q^{2m}e^{-2imz}q^{2m(n - 1)}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{m = 1}^{\infty}q^{2m}e^{2imz}\sum_{n = 1}^{\infty}q^{2m(n - 1)}\right.\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left. - \sum_{m = 1}^{\infty}q^{2m}e^{-2imz}\sum_{n = 1}^{\infty}q^{2m(n - 1)}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{m = 1}^{\infty}\frac{q^{2m}e^{2imz}}{1 - q^{2m}} - \sum_{m = 1}^{\infty}\frac{q^{2m}e^{-2imz}}{1 - q^{2m}}\right)\notag\\ &= \cot z + 4\sum_{n = 1}^{\infty}\frac{q^{2n}}{1 - q^{2n}}\sin 2nz\tag{25}\end{align} Noting that $1 - 2q^{2n}\cos 2z + q^{4n} = (1 - q^{2n}e^{2iz})(1 - q^{2n}e^{-2iz})$ and performing logarithmic differentiation with respect to $q$ we get \begin{align}\frac{1}{\theta_{1}}\frac{\partial \theta_{1}}{\partial q} &= \frac{1}{4q} - \frac{2}{q}\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}} - \frac{2}{q}\left(\sum_{n = 1}^{\infty}\frac{nq^{2n}e^{2iz}}{1 - q^{2n}e^{2iz}} + \sum_{n = 1}^{\infty}\frac{nq^{2n}e^{-2iz}}{1 - q^{2n}e^{-2iz}}\right)\notag\\ &= \frac{1}{4q} - \frac{2}{q}\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}} - \frac{2}{q}\left(\sum_{n = 1}^{\infty}nq^{2n}e^{2iz}\sum_{m = 1}^{\infty}\{q^{2n}e^{2iz}\}^{m - 1}\right.\notag\\ &\left.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \sum_{n = 1}^{\infty}nq^{2n}e^{-2iz}\sum_{m = 1}^{\infty}\{q^{2n}e^{-2iz}\}^{m - 1}\right)\notag\\ &= \frac{1}{4q} - \frac{2}{q}\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}} - \frac{2}{q}\left(\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}nq^{2mn}e^{2imz} + \sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}nq^{2mn}e^{-2imz}\right)\notag\\ &= \frac{1}{4q} - \frac{2}{q}\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}} - \frac{2}{q}\left(\sum_{m = 1}^{\infty}e^{2imz}\sum_{n = 1}^{\infty}nq^{2mn} + \sum_{m = 1}^{\infty}e^{-2imz}\sum_{n = 1}^{\infty}nq^{2mn}\right)\notag\\ &= \frac{1}{4q} - \frac{2}{q}\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}} - \frac{2}{q}\left(\sum_{m = 1}^{\infty}e^{2imz}\frac{q^{2m}}{(1 - q^{2m})^{2}} + \sum_{m = 1}^{\infty}e^{-2imz}\frac{q^{2m}}{(1 - q^{2m})^{2}}\right)\notag\\ &= \frac{1}{4q} - \frac{2}{q}\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}} - \frac{4}{q}\sum_{n = 1}^{\infty}\frac{q^{2n}}{(1 - q^{2n})^{2}}\cos 2nz\notag\end{align} Using differential equation $(24)$ we get $$\frac{1}{\theta_{1}}\frac{\partial^{2} \theta_{1}}{\partial z^{2}} = -1 + 16\sum_{n = 1}^{\infty}\frac{q^{2n}}{(1 - q^{2n})^{2}}\cos 2nz + 8\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}}\tag{26}$$ Now differentiating equation $(25)$ with respect to $z$ we get \begin{align}\frac{\partial }{\partial z}\left(\frac{1}{\theta_{1}}\frac{\partial \theta_{1}}{\partial z}\right) &= -1 - \cot^{2}z + 8\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}}\cos 2nz\notag\\ \Rightarrow \frac{1}{\theta_{1}}\frac{\partial^{2} \theta_{1}}{\partial z^{2}} - \left(\frac{1}{\theta_{1}}\frac{\partial \theta_{1}}{\partial z}\right)^{2} &= -1 - \cot^{2}z + 8\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}}\cos 2nz\notag\\ \Rightarrow \left(\frac{1}{\theta_{1}}\frac{\partial \theta_{1}}{\partial z}\right)^{2} &= 1 + \cot^{2}z - 8\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}}\cos 2nz + \frac{1}{\theta_{1}}\frac{\partial^{2} \theta_{1}}{\partial z^{2}}\notag\end{align} Now using $(25)$ and $(26)$ we get \begin{align}&\left(\cot z + 4\sum_{n = 1}^{\infty}\frac{q^{2n}\sin 2nz}{1 - q^{2n}}\right)^{2}\notag\\ &\,\,\,\,\,\,\,\,= \cot^{2}z + 16\sum_{n = 1}^{\infty}\frac{q^{2n}\cos 2nz}{(1 - q^{2n})^{2}} + 8\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}}(1 - \cos 2nz)\notag\end{align} If we replace $z$ by $\theta/2$ and $q^{2}$ by $x$ and divide resulting equation by $16$ we get the identity $(16)$ obtained by Ramanujan using algebraic manipulation.

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