Modular Equations and Approximations to π(PI): Part 1

In this post we will discuss Ramanujan's classic paper "Modular Equations and Approximations to $\pi$" where Ramanujan offered many amazing formulas and approximations for $\pi$ and showed us the way to create new theories of elliptic and theta functions. However the paper as written in his classic style is devoid of proofs of the most important results. The post would try to elaborate on some of the results mentioned therein.

Class Invariants and Approximations to $\pi$

First and foremost Ramanujan uses his class invariants to find many approximations to $\pi$. From the definitions $$\begin{align}G_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 + e^{-\pi\sqrt{n}})(1 + e^{-3\pi\sqrt{n}})(1 + e^{-5\pi\sqrt{n}})\cdots\notag\\ g_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 - e^{-\pi\sqrt{n}})(1 - e^{-3\pi\sqrt{n}})(1 - e^{-5\pi\sqrt{n}})\cdots\notag\end{align}$$ it is clear that if $n$ is large we can write: $$\begin{align}\pi &\approx \frac{24}{\sqrt{n}}\log(2^{1/4}G_{n})\notag\\ \pi &\approx \frac{24}{\sqrt{n}}\log(2^{1/4}g_{n})\notag\end{align}$$ and clearly the error would be of the order of $e^{-\pi\sqrt{n}}$. Since both $G_{n}$ and $g_{n}$ are algebraic numbers often expressible as surds, the above formulas give approximations to $\pi$ as logarithms of certain surds. Since a modular equation of degree $n$ allows us to calculate $G_{n}$ and $g_{2n}$ (see previous post), it is better to use the formula involving $g_{n}$.

Thus for example if $n = 10$ we have $g_{n} = g_{10} = \sqrt{(1 + \sqrt{5})/2}$ and hence we have the approximation
$$\pi \approx \frac{12}{\sqrt{10}}\log\left(\frac{1 + \sqrt{5}}{\sqrt{2}}\right) = \frac{6}{\sqrt{10}}\log(3 + \sqrt{5}) = 3.14122\cdots$$ so that the approximation is good upto 3 places of decimals and from $g_{18} = (5 + 2\sqrt{6})^{1/6}$ we get $$\pi \approx 4\sqrt{2}\log\{2^{1/4}(5 + 2\sqrt{6})^{1/6}\} = 3.141583\cdots$$ If we use the value of $g_{100}$ we get $$\begin{align}\pi &\approx \frac{24}{\sqrt{100}}\log(2^{1/4}g_{100})\notag\\ &= \frac{12}{5}\log\{2^{-7/8}(1 + \sqrt{5})^{5/4}(3 + 2\sqrt[4]{5})^{1/4}\}\notag\\ &= 3.14159265358973873\ldots\notag\end{align}$$ which is correct to 13 places of decimals.

Ramanujan gives very good approximations using the values of $G_{n}, g_{n}$. Since we have not established the values of these class invariants in our posts we can't do any better than just displaying an approximation of $\pi$ correct to $31$ places of decimals (this is based on the value of $g_{522}$): $$\pi \approx \frac{4}{\sqrt{522}}\log\left[\left(\frac{5 + \sqrt{29}}{\sqrt{2}}\right)^{3}(5\sqrt{29} + 11\sqrt{6})\left\{\sqrt{\frac{9 + 3\sqrt{6}}{4}} + \sqrt{\frac{5 + 3\sqrt{6}}{4}}\right\}^{6}\right]$$ To get to this formula Ramanujan first obtained the value of $g_{58}$ from a modular equation of degree $29$ (that's why we see $\sqrt{29}$ in the above formula) and then used the relation between $g_{9n}$ and $g_{n}$ to calculate $g_{522} = g_{9 \times 58}$.

Ramanujan's function $P(q)$

Next Ramanujan studies a very peculiar function $P(q)$ defined by $$P(q) = 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}}$$ The origin of this function is from the logarithmic derivative of the Dedekind's Eta function $$\eta(q) = q^{1/12}(1 - q^{2})(1 - q^{4})(1 - q^{6})\cdots = q^{1/12}(q^{2};q^{2})_{\infty} = q^{1/12}f(-q^{2})$$ Since $q = e^{-\pi K'/K}$ it follows that $$\frac{dq}{dk} = e^{-\pi K'/K}(-\pi)\frac{d}{dk}\left(\frac{K'}{K}\right) = \frac{\pi^{2}q}{2kk'^{2}K^{2}}$$ (see this post).

And then $$\begin{align}\frac{d}{dk}\{\log(\eta(q))\} &= \frac{dq}{dk}\cdot\frac{d}{dq}\{\log(\eta(q))\}= \frac{dq}{dk}\cdot\frac{d}{dq}\left\{\frac{\log q}{12} + \sum_{n = 1}^{\infty}\log(1 - q^{2n})\right\}\notag\\ &= \frac{dq}{dk}\left\{\frac{1}{12q} - 2\sum_{n = 1}^{\infty}\frac{nq^{2n - 1}}{1 - q^{2n}}\right\}\notag\\ &= \frac{1}{12q}\cdot\frac{dq}{dk}\cdot P(q)\notag\\ &= \frac{\pi^{2}P(q)}{24kk'^{2}K^{2}}\notag\end{align}$$ Now we have $$\begin{align} \eta(q) &= q^{1/12}(q^{2};q^{2})_{\infty} = q^{1/12}f(-q^{2})\notag\\ &= q^{1/12}\sqrt[3]{\phi^{2}(-q^{2})\psi(q^{2})} = q^{1/12}\sqrt[3]{\phi(q)\phi(-q)\psi(q^{2})}\notag\\ &= q^{1/12}\sqrt[3]{2^{-1}q^{-1/4}\theta_{2}(q)\theta_{3}(q)\theta_{4}(q)}\notag\\ &= 2^{-1/3}\theta_{3}(q)\sqrt[3]{\frac{\theta_{2}(q)}{\theta_{3}(q)}\frac{\theta_{4}(q)}{\theta_{3}(q)}}\notag\\ &= 2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\notag\end{align}$$ Let's now suppose that the complete elliptic integrals $L, L'$ for modulus $l$ corresponds to $q^{n}$ so that $$\frac{L'}{L} = n\frac{K'}{K}$$ and $$\frac{dl}{dk} = n\frac{ll'^{2}L^{2}}{kk'^{2}K^{2}}$$ If $n$ is a positive rational then the relation between $k, l$ is an algebraic one and hence $dl/dk$ and $K/L$ are algebraic functions of $k, l$.

Next we can see that $$\frac{\eta(q^{n})}{\eta(q)} = \left(\frac{ll'}{kk'}\right)^{1/6}\sqrt{\frac{L}{K}}\tag{1}$$ and $$\frac{d}{dk}\{\log(\eta(q^{n}))\} = n\frac{\pi^{2}P(q^{n})}{24kk'^{2}K^{2}}$$ so that logarithmic differentiation of the relation betweenn $\eta(q)$ and $\eta(q^{n})$ above yields $$\frac{\pi^{2}}{24kk'^{2}K^{2}}\{nP(q^{n}) - P(q)\} = \frac{1}{6}\frac{d}{dk}\left\{\log\left(\frac{ll'}{kk'}\right)\right\} + \frac{1}{2}\frac{d}{dk}\left\{\log\left(\frac{L}{K}\right)\right\}\tag{2}$$ Since $dl/dk$ and $L/K$ are algebraic functions of $k, l$ it turns out that we can write the above equation in the form $$nP(q^{n}) - P(q) = \frac{KL}{\pi^{2}}\cdot A(k)$$ where $A(k)$ is an algebraic function of $k$ with algebraic numbers as coefficients. If we put $l = k'$ in the above relation we get $q = e^{-\pi/\sqrt{n}}$ and $q^{n} = e^{-\pi\sqrt{n}}$ so that $$nP(e^{-\pi\sqrt{n}}) - P(e^{-\pi/\sqrt{n}}) = \frac{K^{2}}{\pi^{2}}\cdot A\tag{3}$$ where $A$ is an algebraic number dependent on $n$.

Putting $l = k'$ in equation $(1)$ we get $$\frac{\eta(e^{-\pi\sqrt{n}})}{\eta(e^{-\pi/\sqrt{n}})} = n^{-1/4}$$ i.e. $$n^{1/4}e^{-\pi\sqrt{n}/12}(1 - e^{-2\pi\sqrt{n}})(1 - e^{-4\pi\sqrt{n}})\cdots =e^{-\pi/(12\sqrt{n})}(1 - e^{-2\pi/\sqrt{n}})(1 - e^{-4\pi/\sqrt{n}})\cdots$$ Note that this relation is valid for any $n > 0$ and not just for the rational $n$. And hence we can do a logarithmic differentiation with respect to $n$ to get $$nP(e^{-\pi\sqrt{n}}) + P(e^{-\pi/\sqrt{n}}) = \frac{6\sqrt{n}}{\pi}\tag{4}$$ If we put $n = 1$ above we get the beautiful formula $$1 - 24\left(\frac{1}{e^{2\pi} - 1} + \frac{2}{e^{4\pi} - 1}+ \frac{3}{e^{6\pi} - 1} + \cdots\right) = \frac{3}{\pi}$$ Adding equations $(3)$ and $(4)$ we get $$P(e^{-\pi\sqrt{n}}) = \frac{K^{2}}{\pi^{2}}\cdot A + \frac{3}{\pi\sqrt{n}}\tag{5}$$ where $A$ is some algebraic number. Equation $(4)$ is valid for all real numbers $n > 0$, but equations $(3), (5)$ are valid only for rational values of $n$.

Values of $nP(q^{n}) - P(q)$

Ramanujan calculated the values of $nP(q^{n}) - P(q)$ in the form $(4KL/\pi^{2})A(k)$ for many values of $n$ and from his results it is clear that he was interested in the function $$\frac{nP(q^{n}) - P(q)}{\phi^{2}(q^{n})\phi^{2}(q)} = R_{n}(l, k)$$ Here we will do the calculations for $n = 2, 3$. For $n = 2$ we have $l = (1 - k')/(1 + k')$ and hence $$\frac{L}{K} = \frac{1}{1 + l} = \frac{1 + k'}{2}$$ Again $$\begin{align}\left(\frac{ll'}{kk'}\right)^{2} &= \left(\frac{2\sqrt{k'}}{1 + k'}\cdot\frac{1 - k'}{1 + k'}\cdot\frac{1}{kk'}\right)^{2}\notag\\ &= \frac{4(1 - k')^{2}}{k'(1 + k')^{4}(1 - k'^{2})} = \frac{4(1 - k')}{k'(1 + k')^{5}}\notag\end{align}$$ Therefore $$\begin{align}&\frac{1}{6}\frac{d}{dk}\left\{\log\left(\frac{ll'}{kk'}\right)\right\} + \frac{1}{2}\frac{d}{dk}\left\{\log\left(\frac{L}{K}\right)\right\}\notag\\ &\,\,\,\,\,\,\,\,= \frac{1}{12}\frac{d}{dk}\left\{\log\left(\frac{ll'}{kk'}\right)^{2}\right\} + \frac{1}{2}\frac{d}{dk}\left\{\log\left(\frac{L}{K}\right)\right\}\notag\\ &\,\,\,\,\,\,\,\,= \frac{1}{12}\frac{dk'}{dk}\frac{d}{dk'}\left\{\log\left(\frac{4(1 - k')}{k'(1 + k')^{5}}\right)\right\} + \frac{1}{2}\frac{dk'}{dk}\frac{d}{dk'}\left\{\log\left(\frac{1 + k'}{2}\right)\right\}\notag\\ &\,\,\,\,\,\,\,\,= \frac{1}{12}\frac{dk'}{dk}\left(-\frac{1}{k'} - \frac{1}{1 - k'} - \frac{5}{1 + k'} + \frac{6}{1 +k'}\right)\notag\\ &\,\,\,\,\,\,\,\,= \frac{1}{12}\frac{-k}{k'}\left(-\frac{1}{k'} - \frac{1}{1 - k'} + \frac{1}{1 + k'}\right)\notag\\ &\,\,\,\,\,\,\,\,= \frac{1}{12}\cdot\frac{1 + k'^{2}}{kk'^{2}}\notag\end{align}$$ and thus $$\begin{align}2P(q^{2}) - P(q) &= \frac{2K^{2}}{\pi^{2}}(1 + k'^{2})\notag\\ &= \frac{4KL}{\pi^{2}}\cdot\frac{1 + k'^{2}}{1 + k'}\notag\\ &= \frac{4KL}{\pi^{2}}\cdot\frac{1 + l^{2}}{1 + l}\notag\\ &= \frac{4KL}{\pi^{2}}\cdot\frac{1 + l^{2} + k'(1 + l) - 1 + l}{1 + l}\notag\\ &= \frac{4KL}{\pi^{2}}\{k' + l\}\notag\end{align}$$ For $n = 3$ and above it is best to recast the equation $(2)$ in a simpler form. Let us also use the notation $m = K/L$ for Ramanujan's multiplier. Then we have $$\frac{dl}{dk} = \frac{nll'^{2}}{m^{2}kk'^{2}}$$ and $$\begin{align}nP(q^{n}) - P(q) &= \frac{4K^{2}}{\pi^{2}}(kk'^{2})\frac{d}{dk}\left\{\log\left(\frac{ll'}{kk'}\right)\right\} - \frac{12K^{2}}{\pi^{2}}(kk'^{2})\frac{d}{dk}(\log m)\notag\\ &= \frac{4KL}{\pi^{2}}\left(\frac{mkk'^{2}}{ll'}\frac{d}{dk}(ll') - \frac{mkk'^{2}}{kk'}\frac{d}{dk}(kk') - 3kk'^{2}\frac{dm}{dk}\right)\notag\\ &= \frac{4KL}{\pi^{2}}\left(\frac{mkk'^{2}}{ll'}\frac{dl}{dk}\frac{1 - 2l^{2}}{l'} - m(1 - 2k^{2}) - 3kk'^{2}\frac{dm}{dk}\right)\notag\\ \Rightarrow nP(q^{n}) - P(q) &= \frac{4KL}{\pi^{2}}\left((1 - 2l^{2})\frac{n}{m} - (1 - 2k^{2})m - 3kk'^{2}\frac{dm}{dk}\right)\tag{6}\end{align}$$ If $n = 3$ then we have $$\begin{align}l^{2} &= \beta = \frac{(m - 1)^{3}(m + 3)}{16m}\notag\\ k^{2} &= \alpha = \frac{(m - 1)(m + 3)^{3}}{16m^{3}}\notag\\ l'^{2} &= 1 - \beta = \frac{(m + 1)^{3}(3 - m)}{16m}\notag\\ k'^{2} &= 1 - \alpha = \frac{(m + 1)(3 - m)^{3}}{16m^{3}}\notag\end{align}$$ Differentiation yields $$\begin{align} 2k\cdot\frac{dk}{dm} &= \frac{16m^{3}\{(m + 3)^{3} + 3(m - 1)(m + 3)^{2}\} - 48m^{2}(m - 1)(m + 3)^{3}}{256m^{6}}\notag\\ &= \frac{(m + 3)^{2}\{m(m + 3 + 3m - 3) - 3(m - 1)(m + 3)\}}{16m^{4}}\notag\\ &= \frac{(m + 3)^{2}(m - 3)^{2}}{16m^{4}}\notag\end{align}$$ Its time to do some tedious algebra. We have $$\begin{align} \frac{nP(q^{n}) - P(q)}{4KL/\pi^{2}} &= (1 - 2l^{2})\frac{n}{m} - (1 - 2k^{2})m - 3kk'^{2}\frac{dm}{dk}\notag\\ &= \left(1 - \frac{(m - 1)^{3}(m + 3)}{8m}\right)\frac{3}{m} - \left(m - \frac{(m - 1)(m + 3)^{3}}{8m^{2}}\right)\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \frac{96m^{4}}{(m + 3)^{2}(m - 3)^{2}}\cdot k^{2}k'^{2}\notag\\ &= \frac{3 - m^{2}}{m} + \frac{(m - 1)(m + 3)}{8m^{2}}\{(m + 3)^{2} - 3(m - 1)^{2}\}\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \frac{3(m^{2} - 1)(m^{2} - 9)}{8m^{2}}\notag\\ &= \frac{3 - m^{2}}{m} + \frac{(m - 1)(m + 3)}{8m^{2}}\{m^{2} + 6m - 3\}\notag\\ &= \frac{48m - 16m^{3} + (m - 1)(m + 3)(2m^{2} + 12m - 6)}{16m^{2}}\notag\\ &= \frac{48m - 16m^{3}}{16m^{2}}\notag\\ &\,\,\,\,\,\,\,\,\,\, + \frac{(m - 1)(m + 3)(m^{2} + 2m - 3 + m^{2} + 10m - 3)}{16m^{2}}\notag\\ &= \frac{48m - 16m^{3} + (m - 1)^{2}(m + 3)^{2}}{16m^{2}}\notag\\ &\,\,\,\,\,\,\,\,\,\, + \frac{(m^{2} + 2m - 3)(m^{2} + 10m - 3)}{16m^{2}}\notag\\ &= \frac{48m - 16m^{3} + (m - 1)^{2}(m + 3)^{2}}{16m^{2}}\notag\\ &\,\,\,\,\,\,\,\,\,\, + \frac{\{(m + 1)^{2} - 4\}\{(m - 3)^{2} + 16m - 12\}}{16m^{2}}\notag\\ &= \frac{48m - 16m^{3} + (m - 1)^{2}(m + 3)^{2}}{16m^{2}}\notag\\ &\,\,\,\,\,\,\,\,\,\, + \frac{(m + 1)^{2}(m - 3)^{2} + 4(4m^{3} + 4m^{2}- 12m)}{16m^{2}}\notag\\ &= 1 + \frac{(m - 1)^{2}(m + 3)^{2}}{16m^{2}} + \frac{(m + 1)^{2}(m - 3)^{2}}{16m^{2}}\notag\\ &= 1 + kl + k'l'\notag\end{align}$$ So we have $$3P(q^{3}) - P(q) = \frac{4KL}{\pi^{2}}\{1 + kl + k'l'\}$$ From the above it is clear that even if we know the modular equation for some degree $n$, it is very time consuming and tedious to find an expression for $nP(q^{n}) - P(q)$. No one has any idea how Ramanujan calculated the values for $nP(q^{n}) - P(q)$ for large values of $n$. The values of $R_{n}(l, k)$ for $n = 2, 3, 4, 5, 7, 9, 11, 15, 17, 19, 23, 31, 35$ were provided by Ramanujan and we tabulate it below: $$\begin{array}{|r|l|} \hline\\  n & R_{n}(l, k)\\ \hline\\  2 & k' + l\\ \hline\\  3 & 1 + kl + k'l'\\ \hline\\  4 & \dfrac{3(1 + k')(1 + l)}{2}\\ \hline\\  5 & (3 + kl + k'l')\sqrt{\dfrac{1 + kl + k'l'}{2}}\\ \hline\\  7 & 3(1 + kl + k'l')\\ \hline\\  9 & \sqrt{14(1 + k^{2}l^{2} + k'^{2}l'^{2}) + 36(kl + k'l' - kk'll') + \dfrac{96kk'll'}{1 - kl - k'l'}}\\ \hline\\  11 & 2\left\{2(1 + kl + k'l') + \sqrt{kl} + \sqrt{k'l'} - \sqrt{kk'll'}\right\}\\ \hline\\  15 & \left\{1 + (kl)^{1/4} + (k'l')^{1/4}\right\}^{4} - (1 + kl + k'l')\\ \hline\\  17 & \sqrt{A}\\ \hline\\  19 & 6\left\{(1 + kl + k'l') + \sqrt{kl} + \sqrt{k'l'} - \sqrt{kk'll'}\right\}\\ \hline\\  23 & 11(1 + kl + k'l') - 20(4kk'll')^{1/3} - 16(4kk'll')^{1/6}\left(1 + \sqrt{kl} + \sqrt{k'l'}\right)\\ \hline\\  31 & 3B\\ \hline\\  35 & 2\left(\sqrt{kl} + \sqrt{k'l'} - \sqrt{kk'll'}\right) + (4kk'll')^{-1/6}\left(1 - \sqrt{kl} - \sqrt{k'l'}\right)^{3}\\ \hline\end{array}$$ where $$\begin{align} A &= 44(1 + k^{2}l^{2} + k'^{2}l'^{2})\notag\\ &\,\,\,\,\,\, + 168(kl + k'l' - kk'll')\notag\\ &\,\,\,\,\,\, - 102(1 - kl - k'l')(4kk'll')^{1/3}\notag\\ &\,\,\,\,\,\, - 192(4kk'll')^{2/3}\notag\end{align}$$ and $$\begin{align} B &= 3(1 + kl + k'l')\notag\\ &\,\,\,\,\,\, + 4\left(\sqrt{kl} + \sqrt{k'l'} + \sqrt{kk'll'}\right)\notag\\ &\,\,\,\,\,\, - 4(kk'll')^{1/4}\left(1 + (kl)^{1/4} + (k'l')^{1/4}\right)\notag\end{align}$$ The only way these relations have been verified is by using the modular form approach where both the sides are expanded as a series in powers of $q$ and then comparing coefficients upto a certain power of $q$. Under certain conditions if the coefficients in the series of a modular form are zero upto a certain number of terms then the entire modular form is identically zero. However the calculation of the coefficients is performed using some computer algebra packages. So even after almost a century many of the formulas of Ramanujan have not been proved using elementary techniques.

In the same paper Ramanujan says that using the expressions for $nP(q^{n}) - P(q)$ the values of $P(e^{-\pi\sqrt{n}})$ can be calculated through equation $(5)$. And then he just hints that using the values of $rP(q^{r}) - P(q)$ and $sP(q^{s}) - P(q)$ one can calculate $P(e^{-\pi\sqrt{rs}})$. Let $p = rs$ and $k, \gamma, l$ correspond to $q, q^{r} = q', q^{rs} = q^{p} = q'^{s}$ and similarly for the complete elliptic integrals $K, \Gamma, L$. Then we have $$\begin{align}\frac{pP(q^{p}) - P(q)}{\phi^{2}(q^{p})\phi^{2}(q)} &= \frac{pP(q^{p}) - rP(q^{r}) + rP(q^{r}) - P(q)}{\phi^{2}(q^{p})\phi^{2}(q)}\notag\\ &= \frac{r\{sP(q'^{s}) - P(q')\}}{\phi^{2}(q^{p})\phi^{2}(q)} + \frac{rP(q^{r}) - P(q)}{\phi^{2}(q^{p})\phi^{2}(q)}\notag\\ &= \frac{r\{sP(q'^{s}) - P(q')\}}{\phi^{2}(q'^{s})\phi^{2}(q')}\frac{\phi^{2}(q^{r})}{\phi^{2}(q)} + \frac{rP(q^{r}) - P(q)}{\phi^{2}(q^{r})\phi^{2}(q)}\frac{\phi^{2}(q')}{\phi^{2}(q'^{s})}\notag\\ &= rR_{s}(l, \gamma)\frac{\Gamma}{K} + R_{r}(\gamma, k)\frac{\Gamma}{L}\notag\end{align}$$ and therefore $$R_{rs}(l, k) = rR_{s}(l, \gamma)\frac{\Gamma}{K} + R_{r}(\gamma, k)\frac{\Gamma}{L}$$ i.e. $$R_{rs}(l, k) = R_{s}(l, \gamma)\cdot\frac{r}{m_{r}} + R_{r}(\gamma, k)m_{s}\tag{7}$$ where $m_{r} = m_{r}(\gamma, k)$ is multiplier of degree $r$ and $m_{s} = m_{s}(l, \gamma)$ is multiplier of degree $s$.

Again if we put $q = e^{-\pi/\sqrt{n}}$ in equation $(6)$ so that $q^{n} = e^{-\pi\sqrt{n}}$ then we have $l = k', m = \sqrt{n}$ and $$\begin{align} R_{n}(l, k) &= \sqrt{n}(1 - 2l^{2}) - \sqrt{n}(1 - 2k^{2}) - 3kk'^{2}\frac{dm}{dl}\frac{dl}{dk}\notag\\ &= 2\sqrt{n}(k^{2} - l^{2}) - 3ll'^{2}\frac{dm}{dl}\notag\end{align}$$ Changing the notation such that $k$ corresponds to $q = e^{-\pi\sqrt{n}}$ we get $$R_{n}(k, k') = 2\sqrt{n}(1 - 2k^{2}) - 3kk'^{2}\frac{d}{dk}m(k, k')\tag{8}$$ and then clearly the value of $P(e^{-\pi\sqrt{n}})$ is given by $$P(e^{-\pi\sqrt{n}}) = \frac{R_{n}(k, k')}{2\sqrt{n}}\left(\frac{2K}{\pi}\right)^{2} + \frac{3}{\pi\sqrt{n}}\tag{9}$$ Ramanujan used the above value of $P(e^{-\pi\sqrt{n}})$ to give beautiful series for $\pi$. In the next post we will see his ingenuity in deriving these series.

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