Modular Equations and Approximations to π(PI): Part 2

Ramanujan's Series for $ \pi$

Using the values of the function $ P(q)$ for $ q = e^{-\pi\sqrt{n}}$ (see previous post for the definition of $ P(q)$) Ramanujan was able to derive many beautiful series for $ \pi$. He did this in very clever way. The fundamental idea he used was the fact that the function $ \phi^{4}(q) = (2K/\pi)^{2}$ could be expressed in the form of a generalized hypergeometric series.

The General Technique

We first explain the technique in general terms and then give specific examples to illustrate the technique. Let us then suppose that we can express $ (2K/\pi)^{2}$ in the form of a series like $$\left(\frac{2K}{\pi}\right)^{2} = a(k)\sum_{n = 0}^{\infty}b_{n}c^{n}(k)$$ where $ a(k), c(k)$ are algebraic functions of $ k$ and $ b_{n}$ is a rational number. Next we use the fact that $$\eta(q) = q^{1/12}(q^{2};q^{2})_{\infty} = 2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}$$ so that we get $$\eta^{4}(q) = 2^{-4/3}\left(\frac{2K}{\pi}\right)^{2}(kk')^{2/3}$$ and hence we have the relation of the form $$\eta^{4}(q) = d(k)a(k)\sum_{n = 0}^{\infty}b_{n}c^{n}(k)$$ where $ d(k)$ is algebraic function of $ k$. Logarithmic differentiation yields $$\frac{\pi^{2}P(q)}{6kk'^{2}K^{2}} = \frac{d'(k)}{d(k)} + \frac{a'(k)}{a(k)} + \left(\frac{\pi}{2K}\right)^{2}a(k)\sum_{n = 1}^{\infty}nb_{n}c^{n}(k)\frac{c'(k)}{c(k)}$$ so that $$P(q) = \left(\frac{2K}{\pi}\right)^{2}C(k) + D(k)\sum_{n = 1}^{\infty}nb_{n}c^{n}(k)$$ where $ C(k), D(k)$ are algebraic functions of $ k$.

If we put $ q = e^{-\pi\sqrt{n}}$ with $ n$ a positive rational then we see that $$P(e^{-\pi\sqrt{n}}) = \left(\frac{2K}{\pi}\right)^{2}C + D\sum_{m = 1}^{\infty}mb_{m}c^{m}$$ where $ c = c(k), C, D$ are algebraic numbers.

From the previous post we can see that the value of $ P(e^{-\pi\sqrt{n}})$ can be expressed in the form $ (2K/\pi)^{2}A + B/\pi$ with $ A, B \neq 0$ being algebraic numbers. Therefore we get $$\frac{B}{\pi} = \left(\frac{2K}{\pi}\right)^{2}(C - A) + D\sum_{m = 1}^{\infty}mb_{m}c^{m}$$ We can now substitute the value of $ (2K/\pi)^{2}$ from the starting equation and finally get an equation of the form $$\frac{1}{\pi} = \sum_{m = 0}^{\infty}(a + mb)b_{m}c^{m}$$ where $ a, b, c$ are algebraic numbers.

If we do the calculations for $A, B, C, D, a, b, c$ we find that $$\boxed{\displaystyle \begin{align} A &= \frac{R_{n}(k, k')}{2\sqrt{n}},\,\, B = \frac{3}{\sqrt{n}}\notag\\ C &= \frac{3kk'^{2}}{2}\cdot\frac{a'(k)}{a(k)} + (1 - 2k^{2})\notag\\ D &= \frac{3kk'^{2}}{2}\cdot\frac{a(k)c'(k)}{c(k)}\notag\end{align}}$$ $$\boxed{\displaystyle \begin{align} a &= \frac{a(k)(C - A)}{B}\notag\\ &= \frac{\sqrt{n}kk'^{2}}{2}\cdot a'(k) + a(k)\left(\frac{\sqrt{n}}{3}(1 - 2k^{2}) - \frac{R_{n}(k, k')}{6}\right)\notag\\ b &= \frac{D}{B}\notag\\ &= \frac{\sqrt{n}kk'^{2}}{2}\cdot\frac{a(k)c'(k)}{c(k)}\notag\\ c &= c(k)\notag\end{align}}$$ This description fairly sums up Ramanujan's technique. The scheme, although quite non-obvious, when explained looks quite simple, but finding an actual series based on this technique is by no means simple. The skill is in choosing the original series and identifying a suitable function $ c(k)$. The idea is that the value of $ c(k)$ when $ k = k(e^{-\pi\sqrt{n}})$ should be very small when $ n$ is large.

Ramanujan chooses various functions $ c(k)$ which are ultimately tied to the class invariants $ G_{n}, g_{n}$ and provides the final series results and omits the details of the calculations as well as the details of the function $ c(k)$. Even when suitable function $ c(k)$ is chosen the difficult problem is to evaluate $ P(e^{-\pi\sqrt{n}})$ and $ c = c(k(e^{-\pi\sqrt{n}}))$ for a suitably large value of $ n$.

A Simple Example

As an example we will use the following series (see this post): $$\boxed{\displaystyle \left(\frac{2K}{\pi}\right)^{2} = 1 + \left(\frac{1}{2}\right)^{3}(2kk')^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}(2kk')^{4} + \left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3}(2kk')^{6} + \cdots}\tag{1}$$ Here $$ a(k) = 1, b_{n} = (1\cdot 3\cdots (2n - 1))^{3}/(2\cdot 4\cdots (2n))^{3}$$ and $$ c(k) = (2kk')^{2} = G_{n}^{-24}, d(k) = 2^{-4/3}(kk')^{2/3}$$ so that $$ a'(k)/a(k) = 0, c'(k)/c(k) = 2(1 - 2k^{2})/kk'^{2}, d'(k)/d(k) = (2/3)((1 - 2k^{2})/kk'^{2})$$ Thus we get \begin{align} P(q) &= \left(\frac{2K}{\pi}\right)^{2}(1 - 2k^{2}) + 3(1 - 2k^{2})\sum_{n = 1}^{\infty}nb_{n}c^{n}(k)\notag\\ &= (1 - 2k^{2})\sum_{n = 0}^{\infty}(3n + 1)b_{n}c^{n}(k)\tag{2}\end{align} Next we need to calculate $ P(e^{-\pi\sqrt{3}})$. To that end we have (see previous post for details) $$3P(q^{3}) - P(q) = \frac{4KL}{\pi^{2}}\{1 + kl + k'l'\}$$ Here $ l < k$ and if we put $ q = e^{-\pi/\sqrt{3}}$, so that $ l = k', L'/L = \sqrt{3}$ and $$3P(e^{-\pi\sqrt{3}}) - P(e^{-\pi/\sqrt{3}}) = \frac{4L^{2}}{\pi^{2}}\cdot\sqrt{3}\cdot\{1 + 2kk'\}$$ Since we have $ \sqrt{kl} + \sqrt{k'l'} = 1$ (modular equation of degree $ 3$), it follows by putting $ l = k'$ that $ 2kk' = 1/2$ and therefore $$3P(e^{-\pi\sqrt{3}}) - P(e^{-\pi/\sqrt{3}}) = \frac{L^{2}}{\pi^{2}}\cdot 6\sqrt{3}$$ And we have $$3P(e^{-\pi\sqrt{3}}) + P(e^{-\pi/\sqrt{3}}) = \frac{6\sqrt{3}}{\pi}$$ Adding these equations we get $$P(e^{-\pi\sqrt{3}}) = \sqrt{3}\left(\frac{L^{2}}{\pi^{2}} + \frac{1}{\pi}\right)$$ Here we need to understand that $ L = K(k(e^{-\pi\sqrt{3}}))$ and hence if we change the notation so that $ q = e^{-\pi\sqrt{3}}, k = k(q)$ then we can write $$P(q) = \sqrt{3}\left(\frac{K^{2}}{\pi^{2}} + \frac{1}{\pi}\right)\tag{3}$$ Now we need to calculate the value of $ k$. Clearly from the modular equation of degree $ 3$ we see that $ kk' = 1/4$ and we must have $ k < k'$. Thus we have \begin{align}&k^{2}(1 - k^{2}) = \frac{1}{16}\notag\\ &\Rightarrow \alpha - \alpha^{2} = \frac{1}{16}\notag\\ &\Rightarrow 16\alpha^{2} - 16\alpha + 1 = 0\notag\\ &\Rightarrow \alpha = \frac{16 - \sqrt{256 - 64}}{32}\text{ (we choose smaller root here)}\notag\\ &\Rightarrow k^{2} = \frac{2 - \sqrt{3}}{4}\notag\end{align} From equations $ (2)$ and $ (3)$ (and using the value of $ k^{2}$ above) we get \begin{align}&\sqrt{3}\left(\frac{K^{2}}{\pi^{2}} + \frac{1}{\pi}\right) = \frac{\sqrt{3}}{2}\sum_{n = 0}^{\infty}(3n + 1)b_{n}c^{n}(k)\notag\\ &\Rightarrow \frac{4K^{2}}{\pi^{2}} + \frac{4}{\pi} = \sum_{n = 0}^{\infty}(6n + 2)b_{n}c^{n}(k)\notag\\ &\Rightarrow \sum_{n = 0}^{\infty}b_{n}c^{n}(k) + \frac{4}{\pi} = \sum_{n = 0}^{\infty}(6n + 2)b_{n}c^{n}(k)\notag\\ &\Rightarrow \frac{4}{\pi} = \sum_{n = 0}^{\infty}(6n + 1)b_{n}c^{n}(k)\notag\end{align} Since $ c(k) = (2kk')^{2} = 1/4$ it follows that we have $$ \boxed{\displaystyle \frac{4}{\pi} = 1 + \frac{7}{4}\left(\frac{1}{2}\right)^{3} + \frac{13}{4^{2}}\left(\frac{1\cdot 3}{2\cdot 4}\right)^{3} + \frac{19}{4^{3}}\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3} + \cdots}\tag{4}$$ This is the first and the simplest series Ramanujan obtains with this technique. Based on the same function $ c(k)$ Ramanujan gives further series corresponding to modular equations of degree $ n = 7, 15$.

For general $ n$ it is easily seen that the value of $ P(e^{-\pi\sqrt{n}})$ is given by $$P(e^{-\pi\sqrt{n}}) = \frac{R_{n}(k, k')}{2\sqrt{n}}\left(\frac{2K}{\pi}\right)^{2} + \frac{3}{\pi\sqrt{n}}$$ Using this result and noting that $ 2kk' = G_{n}^{-12}$ we can write the general formula as follows: $$\boxed{\displaystyle \frac{1}{\pi} = \sum_{m = 0}^{\infty}\left(\frac{1\cdot 3\cdot 5\cdots (2m - 1)}{2\cdot 4\cdot 6\cdots (2m)}\right)^{3}(a + bm)G_{n}^{-24m}}$$ where $$a = \frac{\sqrt{n}}{3}(1 - 2k^{2}) - \frac{R_{n}(k, k')}{6},\,\,\,b = \sqrt{n}(1 - 2k^{2})$$ In the formula $(1)$ we note that $K, k, k'$ are functions of nome $q$ and hence replacing $q$ by $-q$ we get: \begin{align}\left(\frac{2K}{\pi}\right)^{2} = \theta_{3}^{4}(q) &\rightarrow \theta_{3}^{4}(-q)\notag\\ &= \theta_{4}^{4}(q) = \frac{\theta_{4}^{4}(q)}{\theta_{3}^{4}(q)}\theta_{3}^{4}(q)\notag\\ &= k'^{2}\left(\frac{2K}{\pi}\right)^{2}\notag\end{align} \begin{align}(2kk')^{2} &= 4k^{2}k'^{2} = 4(1 - k'^{2})k'^{2}\notag\\ &= 4\left(1 - \frac{\theta_{4}^{4}(q)}{\theta_{3}^{4}(q)}\right)\frac{\theta_{4}^{4}(q)}{\theta_{3}^{4}(q)}\notag\\ &\rightarrow 4\left(1 - \frac{\theta_{4}^{4}(-q)}{\theta_{3}^{4}(-q)}\right)\frac{\theta_{4}^{4}(-q)}{\theta_{3}^{4}(-q)}\notag\\ &= 4\left(1 - \frac{\theta_{3}^{4}(q)}{\theta_{4}^{4}(q)}\right)\frac{\theta_{3}^{4}(q)}{\theta_{4}^{4}(q)}\notag\\ &= 4\left(1 - \frac{1}{k'^{2}}\right)\frac{1}{k'^{2}}\notag\\ &= \frac{-4k^{2}}{k'^{4}} = -\left(\frac{2k}{k'^{2}}\right)^{2} = -g_{n}^{-24}\notag\end{align} Hence we get another formula $$\boxed{\displaystyle \left(\frac{2K}{\pi}\right)^{2} = \frac{1}{k'^{2}}\sum_{m = 0}^{\infty}(-1)^{m}\left(\frac{1\cdot 3 \cdot 5\cdots (2m - 1)}{2 \cdot 4 \cdot 6\cdots (2m)}\right)^{3}g_{n}^{-24m}}\tag{5}$$ If we apply Ramanujan's technique on the above formula we obtain the following: $$\boxed{\displaystyle \frac{1}{\pi} = \sum_{m = 0}^{\infty}(-1)^{m}\left(\frac{1\cdot 3 \cdot 5\cdots (2m - 1)}{2 \cdot 4 \cdot 6\cdots (2m)}\right)^{3}(a + bm)g_{n}^{-24m}}$$ where $$a = \frac{\sqrt{n}(1 + k^{2})}{3k'^{2}} - \frac{R_{n}(k, k')}{6k'^{2}},\,\,\, b = \frac{\sqrt{n}(1 + k^{2})}{1 - k^{2}}$$ Putting $n = 2$ and noting that $k = (\sqrt{2} - 1), g_{2} = 1, R_{2}(k, k') = 2k$ we get another formula of Ramanujan: $$\boxed{\displaystyle \frac{2}{\pi} = \sum_{m = 0}^{\infty}(-1)^{m}\left(\frac{1\cdot 3 \cdot 5\cdots (2m - 1)}{2 \cdot 4 \cdot 6\cdots (2m)}\right)^{3}(4m + 1)}$$ If we replace $k$ by $(1 - k')/(1 + k')$ in $(5)$ and note that $k^{2}/2k' = g_{4n}^{-12}$ then we get: $$\boxed{\displaystyle \left(\frac{2K}{\pi}\right)^{2} = \frac{1}{k'}\sum_{m = 0}^{\infty}(-1)^{m}\left(\frac{1\cdot 3 \cdot 5\cdots (2m - 1)}{2 \cdot 4 \cdot 6\cdots (2m)}\right)^{3}g_{4n}^{-24m}}\tag{6}$$ and then we get another general series for $1/\pi$: $$\boxed{\displaystyle \frac{1}{\pi} = \sum_{m = 0}^{\infty}(-1)^{m}\left(\frac{1\cdot 3 \cdot 5\cdots (2m - 1)}{2 \cdot 4 \cdot 6\cdots (2m)}\right)^{3}(a + bm)g_{4n}^{-24m}}$$ where \begin{align}a &= \sqrt{n}g_{4n}^{-12} + \frac{1}{k'}\left(\frac{\sqrt{n}}{3}(1 - 2k^{2}) - \frac{R_{n}(k, k')}{6}\right)\notag\\ b &= \sqrt{n}\left(k' + \frac{1}{k'}\right)\notag\end{align} In the next post we will discuss the series based on alternative theories developed by Ramanujan.

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