6
$\begingroup$

In his classic paper "Modular Equations and Approximations to $\pi$ (1914)", Ramanujan gives a standard technique to obtain a general family of series for $1/\pi$ based on series for $(2K/\pi)^{2}$ in terms of $k$ (see the details here). Most of the series which Ramanujan provides are based on the following expressions for $(2K/\pi)^{2}$:

\begin{align}\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} + \cdots\tag{1}\\ \left(\frac{2K}{\pi}\right)^{2}&= (1 + k^{2})^{-1}\,_{3}F_{2}\left(\frac{1}{4},\frac{3}{4},\frac{1}{2}; 1, 1; \left(\frac{g^{12} + g^{-12}}{2}\right)^{-2}\right)\tag{2}\\ \left(\frac{2K}{\pi}\right)^{2}&= (1 - 2k^{2})^{-1}\,_{3}F_{2}\left(\frac{1}{4},\frac{3}{4},\frac{1}{2}; 1, 1; -\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2}\right)\tag{3}\\ \left(\frac{2K}{\pi}\right)^{2}&= \{1 - (kk')^{2}\}^{-1/2}\,_{3}F_{2}\left(\frac{1}{6},\frac{5}{6}, \frac{1}{2};1;1; \frac{27G^{24}}{(4G^{24} - 1)^{3}}\right)\tag{4}\end{align}

(In the above $G = (2kk')^{-1/12}, g = (2k/k'^{2})^{-1/12}$. I have left the series related to functions ${}_{3}F_{2}(1/3, 2/3, 1/2; 1; 1; a(k))$ based on elliptic functions to base 3).

Next Chudnovsky brothers (around 1989) use the following series $$\left(\frac{2K}{\pi}\right)^{2} = \{1 - 4G^{-24}\}^{-1/2}\,_{3}F_{2}\left(\frac{1}{6},\frac{5}{6}, \frac{1}{2};1;1; \frac{-27G^{48}}{(G^{24} - 4)^{3}}\right)\tag{5}$$ to give their famous series based on $G_{163}$. Another series can be obtained from $(5)$ above by changing the nome $q$ into $(-q)$: $$\left(\frac{2K}{\pi}\right)^{2} = \{k'^{4} + 16k^{2}\}^{-1/2}\,_{3}F_{2}\left(\frac{1}{6},\frac{5}{6}, \frac{1}{2};1;1; \frac{27g^{48}}{(g^{24} + 4)^{3}}\right)\tag{6}$$

The general family of series for $1/\pi$ based on equation $(6)$ is as follows: $$\frac{1}{\pi} = \sum_{m = 0}^{\infty}\frac{(1/6)_{m}(5/6)_{m}(1/2)_{m}}{(m!)^{3}}(A + mB)X_{n}^{m}\tag{7}$$ where \begin{align} X_{n} &= \frac{27g_{n}^{48}}{(g_{n}^{24} + 4)^{3}}\notag\\ A &= \frac{1}{2k\sqrt{g_{n}^{24} + 4}}\left(\frac{\sqrt{n}}{3}(1 - 2k^{2}) - \frac{R_{n}(k, k')}{6}\right) - \sqrt{n}\cdot\frac{g_{n}^{12}(k^{2} + 7)}{4(g_{n}^{24} + 4)^{3/2}}\notag\\ B &= \sqrt{n}(g_{n}^{12} + k)\cdot\frac{g_{n}^{24} - 8}{(g_{n}^{24} + 4)^{3/2}}\notag\\ n &> 4\notag\\ k &= k(q) = k(e^{-\pi\sqrt{n}})\notag \end{align}

I am currently trying to work out a series based on the above equation $(7)$ (using $n = 58$) and bit bogged down into calculations with various radicals. I don't know if the equation $(7)$ above has been used anywhere to generate series for $1/\pi$ (let me know of any references if this has already been in literature). I want to know whether this approach would lead a genuine new family of series for $1/\pi$.

$\endgroup$
4
$\begingroup$

See page 5 in http://arxiv.org/abs/1112.3259 Are the series presented there somewhat related to your series (7)?

$\endgroup$
  • 1
    $\begingroup$ This paper gives tranformations to obtain a series for $1/\pi$ from a given series for $1/\pi$. While the series given in the paper are not exactly the same as I have derived equation $(7)$, I must give +1 for the nice paper. $\endgroup$ – Paramanand Singh Feb 24 '15 at 3:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.