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In his famous paper "Modular Equations and Approximations to $\pi$", Ramanujan gives the following famous series for $1/\pi$: \begin{align}\frac{1}{2\pi\sqrt{2}} &= \frac{1103}{99^{2}} + \frac{27493}{99^{6}}\frac{1}{2}\frac{1\cdot 3}{4^{2}} + \frac{53883}{99^{10}}\frac{1\cdot 3}{2\cdot 4}\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}} + \cdots\notag\\ &= \sum_{n = 0}^{\infty}\dfrac{\left(\dfrac{1}{4}\right)_{n}\left(\dfrac{1}{2}\right)_{n}\left(\dfrac{3}{4}\right)_{n}}{(n!)^{3}}(1103 + 26390n)\left(\frac{1}{99^{2}}\right)^{2n + 1}\notag\end{align}

He also mentions the technique for finding such series which is based on the evaluation of $nP(q^{n}) - P(q)$ in a closed form. Here $$P(q) = 1 - 24\sum_{j = 1}^{\infty}\frac{jq^{2j}}{1 - q^{2j}}$$ The kind of closed form needed is $$nP(q^{n}) - P(q) = \frac{4LK}{\pi^{2}}\cdot A(l, k)$$ where $k, l$ and $K, L$ correspond to $q, q^{n}$ and $A(l, k)$ is an algebraic function. In order to derive the series mentioned above it is necessary to calculate this expression $nP(q^{n}) - P(q)$ for $n = 58 = 2\cdot 29$ which can be done (as mentioned by Ramanujan) if we can calculate its value for $n = 2$ and $n = 29$. Sadly Ramanujan does not give the expression $A(l, k)$ for $n = 29$. I consulted books of Bruce C. Berndt but could not find this specific expression. Although Ramanujan mentions a process where this expression can be obtained from a modular equation of degree $29$, but due to the complexity of Russell's modular equation of degree $29$ I can't apply the technique.

Is there any work (paper) available which tries to directly use Ramanujan's approach and prove the above series by calculating $A(l, k)$ for $n = 29$ from a modular equation?

Note: There does not seem to be (Edit: this has changed since then) a specific tag related to Ramanujan so I have put this under "sequences-and-series" and noting that nowadays most of Ramanujan's work is studied under modular-forms I have added that tag.

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  • $\begingroup$ See cecm.sfu.ca/~pborwein/PAPERS/P40.pdf. $\endgroup$ – Dietrich Burde Apr 20 '14 at 7:55
  • $\begingroup$ @DietrichBurde: I read that paper and found that regarding this series Borwein mentions about calculation of $\alpha(58)$ which leads to $1103$ in the series. He writes "It is less clear how one calculates $\alpha(58)$ in algebraic form... but a numerical calculation ...is easily obtained" and this confirms the value $1103$. Borwein believes that "this is presumably what Ramanujan observed". Then from the number $1103$ and value of $g_{58}$ the value of $\alpha(58)$ in algebraic form is obtained. $\endgroup$ – Paramanand Singh Apr 20 '14 at 8:12
  • $\begingroup$ @DietrichBurde: Continuing from prev comment. Borwein then says that the agreement of the sum of series with $1/\pi$ to 3 millions places confirms that $\alpha(58)$ should have a specific algebraic value. It then seems that no one has really calculated the value of $\alpha(58)$ directly by manipulating radicals/solving equation etc, but all is based on numerical calculations and then some knowledge about algebraic nature of $\alpha(58)$. I strongly believe that Ramanujan did calculate the value $1103$ exactly and not a guess based on numerical calculations. $\endgroup$ – Paramanand Singh Apr 20 '14 at 8:15
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I would like to outline a proof of this famous identity, which is closely related to the question I have posted on MathOverflow. It is somewhat different from the proof of Borwein brothers.

Addendum: I can use a construction given by Mazur-Swinnerton-Dyer and Zagier to prove Ramanujan's identity for $n=37$.

Let $n=58$.

1. Following Borwein brothers, we can get

$$\frac{1}{\pi}=\sum_{m=0}^{\infty}(2\sqrt{n}v(k)m+G_0)b_mc^m(k)$$

where

$$b_m=\frac{(4m)!}{4^{4m}(m!)^4}$$

$$2v(k)=\left(1-\frac{2}{((k^{\prime})^2/(2k))^2+1}\right)$$

$$c(k)=\left(\frac{2}{2k/(k^{\prime})^2+(k^{\prime})^2/(2k)}\right)^2$$

$$G_0=\frac{\sqrt{n}}{3}\left(1-\frac{3}{2(((k^{\prime})^2/(2k))^2+1)}-\frac{1}{1+k^2}\frac{G_1}{2}\right)$$

$$G_1=\frac{nP(q^n)-P(q)}{(2K(k)/\pi)^2}$$

2. Following H. M. Weber, one has

$$\frac{2k}{(k^{\prime})^2}=\left(\frac{\sqrt{29}-5}{2}\right)^6$$

where $k=k(e^{-\pi\sqrt{58}})$

3. $$\frac{G_1}{1+k^2}=\frac{nP(q^n)-P(q)}{\eta^{2}(q^{2n})\eta^{2}(q^{4})}\frac{c(k)^{1/4}}{8\sqrt{n}}$$

Denote $$H(q)=\frac{nP(q^n)-P(q)}{\eta^{2}(q^{2n})\eta^{2}(q^{4})}$$

Then $H(q)^2$ is a weakly modular form on $\mathbb{H}/\Gamma_0(58)$

4. Denote $$[a_1,\cdots,a_n]=\prod_{\delta\mid N,\sum a_\delta=0}\eta^{a_\delta}(\delta\tau)$$

where $\eta$ is Dedekind eta function. Then $H(q)^2\cdot[-2,8,10,-16]$ is holomorphic on $\mathbb{H}/\Gamma_0(58)$ except at infinity. Then it is a linear combination of eta function product invariant under $\Gamma_0(58)$.

5. $[\cdots,\alpha_\delta,\cdots]$ is invariant under $\Gamma_0(58)$ if 1)$24\mid\sum_{\delta\mid 58}\delta a_{\delta}$; 2) $24\mid\sum_{\delta\mid 58}58a_{\delta}/\delta $ ; 3)$\prod_{\delta \mid n}\delta^{a_\delta}$ is a rational square. What's more, eta product is holomorphic at a cusp $c/d$ if

$$\frac{1}{24}\sum_{\delta\mid 58}\frac{(\mathrm{gcd}(d,\delta))^2}{\delta}a_\delta\geq 0$$

6. Expand $H(q)^2\cdot[-2,8,10,-16]$ and eta products, use matkerint function in PARI/GP to calculate the coefficients of linear combination. Note that

$$[a,b,c,d]=2^{c/2}58^{(a+b)/4}\left(\frac{\sqrt{2}}{2}\frac{\sqrt{29}+5}{2}\right)^{(a+d)/2}$$

where $q=\exp(-\pi/\sqrt{58})$. All these would lead to

$$G_0=\frac{\sqrt{58}}{3}\left(1-\frac{3}{4\times 99^2}\left(\frac{\sqrt{29}-5}{2}\right)^6-\frac{36\sqrt{2}(148 + 11 \sqrt{29})}{99\times16\sqrt{58}}\right)$$

and we are done.

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    $\begingroup$ This is much beyond my understanding and I need to study a lot more stuff to figure all this out in detail. Also I am not able to understand the definition of $[a, b, c, d]$ because the defining equation does not have $a, b, c, d$ on right side. Also $\delta$ is supposed to be a positive divisor of $N = 58$, but no idea as to what is $a_{\delta}$. Some more information on this would definitely help. +1 by the way to get the desired value of the constant $G_{0}$. $\endgroup$ – Paramanand Singh Mar 12 '15 at 4:10
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    $\begingroup$ @ParamanandSingh: I have corrected some typos and rewritten the definition of $[\cdots, a_\delta, \cdots]$. More details would be added in the near future. $\endgroup$ – Y. Zhao Mar 12 '15 at 9:09
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See: J.M Borwein, P.B. Borwein, "Pi and AGM". John Wiley and Sons, Inc. New York, Chichester, Brisbane, Toronto, Singapore, 1987.

pages 172, 177.

I have a note and a Berndt's formula

1) Let $$ A_{p,r}:=\frac{f(-q^2)}{q^cf(-q^{2p})}\textrm{, }c=\frac{p-1}{12} $$ then $T_{p,r}=P(q)-pP(q^p)$ is $$ T_{p,r}=\frac{24\sqrt{r}}{\pi A_{p,r}}\frac{dA_{p,r}}{dr}. $$ 2) The formula in Berndt's book [1] is $$ T_{p,r}=\frac{16}{\pi^2}(k_rk'_r)^2K^2\frac{d}{dk}\log\left(m_{p}(r)^{3}\frac{k_rk'_r}{k_{p^2r}k'_{p^2r}}\right) $$

[1]: B.C. Berndt. 'Ramanujan's Notedbooks Part III'. 1991 ed., New York: Springer-Verlag.

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  • $\begingroup$ I have that book and unfortunately it does not provide the details of the calculation for this specific series of Ramanujan. $\endgroup$ – Paramanand Singh Mar 12 '16 at 13:27
  • $\begingroup$ Have you try to use the formula $\endgroup$ – Nikos Bagis Oct 28 '16 at 20:44
  • $\begingroup$ I had tried to obtain the expression for $m$ by differentiating the modular equation of degree 29. But this lead to very complex expressions and again differentiating it was simply not possible for me. Is there an expression for multiplier $m$ for degree 29 available in literature? $\endgroup$ – Paramanand Singh Oct 29 '16 at 4:01

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