Ramanujan's series for $(1/\pi)$ and modular equation of degree $29$

Question

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.

Answer 1

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.

Answer 2

In https://arxiv.org/abs/1911.03968, I provide a complete proof of the Ramanujan's series for $1/\pi$ in the question. First, we obtain, using a Maple procedure, a modular equation of level $2$ and degree $29$ in the Russell form: $$ u^2=\alpha \beta, \quad v^2=(1-\alpha)(1-\beta), \quad P(u,v)=0, $$ where $P(u,v)$ is a long symmetric polynomial of degree $15$ in both variables $u$ and $v$ (For readers that do not use Maple, we show the modular equation in the Appendix). From it, we get the proof of that impresive series for $1/\pi$ due to Ramanujan.

Answer 3

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.

Answer 4

I analyzed the series in question as well as the one given by Chudnovsky brothers and also the following series given by Ramanujan $$\frac{4}{\pi}=\sum_{n=0}^{\infty} (-1)^n\frac{(1/4)_n(2/4)_n(3/4)_n}{(n!)^3}(1123+21460n)\left(\frac{1}{882}\right)^{2n+1}$$ All these series which can be obtained from Ramanujan's technique can be expressed in the format $$\frac{1}{\pi}=\sum_{n=0}^{\infty} (a_N+nb_N)d_n (c_N) ^{n} $$ where $N$ is a positive integer and $a_N, b_N, c_N $ are algebraic numbers dependent on $N$ and are based on modular equations of degree $N$. Further $d_n$ is a typical sequence involving ratio of rising factorials.

In each case it can be proved that $a_N\sqrt{N} /b_N\to1/\pi$ as $N\to\infty $. This is quite obvious if one studies the formulation given by Chudnovsky brothers. For the technique provided by Ramanujan it needs some patience to verify this. This is also related to the fact that the function $\alpha(N) $ given by Borwein brothers in Pi and the AGM tends to $1/\pi$.

By a more careful analysis of the rate of convergence of the ratio $a_N\sqrt {N} /b_N$ it is possible to evaluate $a_N$ given the value $b_N$.

One can easily find a suitable rational approximation $p/q$ for $1/\pi\sqrt {N} $ and then just use $a_N=pb_N/q$.

This is so evident in the approximations $$\frac{1123}{21460}\approx\frac{1}{\pi\sqrt{37}}$$ and $$\frac{1103}{26390}\approx\frac{1}{\pi\sqrt{58}}$$ and $$\frac{13591409}{545140134} \approx\frac{1}{\pi\sqrt{163}}$$ Even for small values of $N$ eg $N=7$ the following approximation $1/\pi\sqrt{7}\approx 5/42$ works fine to give the series $$\frac{1}{\pi}=\sum_{n=0}^{\infty} \binom{2n}{n}^3\frac{42n+5}{2^{12n+4}}$$ (no other integer $p$ except $5$ gives a better approximation of the form $p/42$ for $1/\pi\sqrt{7}$).