A formula in Ramanujan's lost notebook and its connection with Chudnovsky series for $1/\pi$

Question

While studying Berndt's Ramanujan's Lost Notebook Vol. 2, page 369 (chapter on Springerlink), I found that Ramanujan gave values of a certain expression $$\frac{1}{\sqrt{Q_n}}\left(\sqrt {n} P_n-\frac{6}{\pi}\right)\tag{1}$$ for a few integer values of $n$. Most notable among them is the value for $n=163$ with $$\frac{1}{\sqrt{Q_n}}\left(\sqrt {n} P_n-\frac{6}{\pi}\right)=362\sqrt{\frac{3}{3335}}\tag{2}$$ Let's now define the symbols used above to get full context needed for the question. We have by definition \begin{align} P(q) & =1-24\sum_{j=1}^{\infty} \frac{jq^j} {1-q^j}\tag{3a}\\ Q(q) & =1+240\sum_{j=1}^{\infty} \frac{j^3q^j} {1-q^j}\tag{3b}\\ R(q) & =1-504\sum_{j=1}^{\infty} \frac{j^5q^j} {1-q^j}\tag{3c} \end{align} and $$P_n=P(-e^{-\pi\sqrt {n}}), Q_n=Q(-e^{-\pi\sqrt {n}}), R_n=R(-e^{-\pi\sqrt{n}})\tag{4}$$ These expressions famously appear in the general series for $1/\pi$ given by Chudnovsky brothers $$ \frac{1}{\pi} = \frac{1}{\sqrt{-j_{n}}}\sum_{m = 0}^{\infty}\frac{(6m)!}{(3m)!(m!)^{3}}\frac{a_{n} + mb_{n}}{j_{n}^{m}}\tag{5}$$ where \begin{align} j_{n} &= 1728\frac{Q_{n}^{3}}{Q_{n}^{3} - R_{n}^{2}}\tag{6a}\\ b_{n} &= \sqrt{n(1728 - j_{n})}\tag{6b}\\ a_{n} &= \frac{b_{n}}{6}\left\{1 - \frac{Q_{n}}{R_{n}}\left(P_{n} - \frac{6}{\pi\sqrt{n}}\right)\right\}\tag{6c} \end{align} We can see that Ramanujan's expression $(1)$ is related to $a_n$ in $(6c)$.

Berndt obtained the value in $(2)$ by using a table of values of $a_n, b_n$. Berndt's book does not indicate the procedure to obtain these numbers.

My question is whether Ramanujan had some inkling of the Chudnovsky series or not. At least the available literature does not give any explicit details regarding this. Also it is not clear how he computed the value of expression $(1)$ for $n=163$. Is there any way to evaluate this without using $a_n, b_n$?

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It is interesting to observe however that the Chudnovsky series $(5)$ can be obtained using the approach described by Ramanujan in his 1914 paper Modular Equations and Approximations to $\pi$ (Link).

Update: In the above paper Ramanujan gives a technique to evaluate $P(q^2)$ in closed form for $q=e^{-\pi\sqrt{n}} $ for many integer values of $n$. Using the identity $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)$$ one can then calculate the value of $P(-q) $ in closed form. I tried to do the calculations for $n=11$ but it seems too formidable and I couldn't complete it so far. Also the paper by Ramanujan does not have the formulas related to $n=43,67,163$ and thus it appears that he did develop some formulas for these values after his 1914 paper, but somehow failed to record them in his lost notebook and instead gave direct evaluation of expression $(1)$ for some values of $n$.

Answer 1

Ramanujan gave the value $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n}P_n-\frac{6}{\pi}\right)=\sqrt{2}\tag{1}$$ for $n=11$ and I have now completed the calculations needed to prove this using an approach suggested at the end of the question.

We need two ingredients here with the first one being the modular equation of degree $11$ given by Ramanujan $$\sqrt{kl} +\sqrt{k'l'} +2(4klk'l')^{1/6}=1\tag{1}$$ The equation is symmetric in $k, l$ and hence works when modulus $l$ is of degree $11$ over modulus $k$ as well as when $k$ is of degree $11$ over $l$. Let us assume the latter here and let $k, l$ correspond to nomes $e^{-\pi\sqrt{11}},e^{-\pi/\sqrt{11}} $ respectively so that $l=k'$ and then the equation $(1)$ reduces to $$2\sqrt{kk'}+2(2kk')^{1/3}=1$$ Let us further observe that $G=G_{11}=(2kk')^{-1/12}$ and hence the above equation can be written in terms of $G$ as $$2(G^{-12}/2)^{1/2}+2G^{-4}=1$$ or $$\sqrt{2}+2G^{2}=G^6$$ so that $a=G^2=G_{11}^2$ is the root of $$f(x) = x^3-2x-\sqrt{2}\tag{2}$$ It can be checked that this equation has no roots in $\mathbb{Q} (\sqrt{2})$ and hence $a$ is of degree $3$ over $\mathbb{Q} (\sqrt{2})$ and hence of degree $6$ over $\mathbb{Q} $.

The minimal polynomial for $a$ over rationals is then $$(x^3-2x-\sqrt{2})(x^3-2x+\sqrt{2})$$ or $$x^6-4x^4+4x^2-2$$ so that $b=a^2=G_{11}^4$ is a root of $$g(x) =x^3-4x^2+4x-2\tag{3}$$ Next ingredient we need is the expression for $nP(q^{2n})-P(q^2)$ for $n=11$ provided by Ramanujan as $$nP(q^{2n})-P(q^2)=\frac{8KL}{\pi^2}\left\{2(1+kl+k'l')+\sqrt{kl} +\sqrt{k'l'} - \sqrt{klk'l'} \right\}\tag{4}$$ and here we can again assume $k$ being of degree $11$ over $l$ so that $l$ corresponds to $q$ and $k$ to $q^{n} $ and $K'/K=nL'/L$. Putting $q=e^{-\pi/\sqrt{11}}$ so that $l=k'$ and $L/K=K'/K=\sqrt {11}$ we get $$nP(q^{2n})-P(q^2)=\sqrt{11}\left(\frac{2K}{\pi}\right)^2(4(1+2kk')+4\sqrt{kk'}-2kk')$$ or $$nP(q^{2n})-P(q^2)=\sqrt{11}\left(\frac{2K}{\pi}\right)^2(4(1+G^{-12})+2\sqrt{2}G^{-6}-G^{-12})$$ or $$nP(q^{2n})-P(q^2)=\sqrt{11}\left(\frac{2K}{\pi}\right)^2(4+3b^{-3}+2\sqrt{2}a^{-3})$$ Using $\sqrt{2}=a^3-2a$ we get $$nP(q^{2n})-P(q^2)=\sqrt{11}\left(\frac{2K}{\pi}\right)^2\frac{6b^3-4b^2+3}{b^3}$$ We need another identity $$nP(q^{2n})+P(q^2)=\frac{6\sqrt{n}}{\pi}$$ which holds for $q=e^{-\pi/\sqrt{n}} $. Using this identity together with previous equation we get $$2P(e^{-2\pi\sqrt {11}})=\frac{6}{\pi\sqrt{11}}+\frac{1}{\sqrt{11}} \left(\frac{2K}{\pi}\right)^2\frac{6b^3-4b^2+3}{b^3}\tag{5}$$ Next we use the identity $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)$$ with $q=e^{-\pi\sqrt{11}}$ to get $$P(-q)=\frac{6}{\pi\sqrt{11}}+\left(\frac {2K}{\pi}\right)^2\left(\frac{6b^3-4b^2+3}{b^3\sqrt{11}}-(1-2k^2)\right)$$ or $$\sqrt{n} P_n-\frac{6}{\pi}=\left(\frac {2K}{\pi}\right)^2\left(\frac{6b^3-4b^2+3}{b^3}-\frac{\sqrt{11(b^6-1)}}{b^3}\right)\tag{5}$$ as $$(1-2k^2)^2=1-G^{-24}=1-b^{-6}$$ To evaluate $Q_n$ we note that $$Q(-q) =\left(\frac{2K}{\pi}\right) ^4(1-4G^{-24})=\left(\frac{2K}{\pi}\right) ^4\frac{b^6-4}{b^6}$$ and hence $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n}P_n-\frac{6}{\pi}\right)=\frac{6b^3-4b^2+3-\sqrt{11(b^6-1)}}{\sqrt {b^6-4}}$$ The expression on right simplifies to $\sqrt{2}$ in a miraculous manner as demonstrated in this answer of mine.

A similar calculation can be done for $n=19$ by noting that $2^{1/4}G_{19}$ is the real root of $x^3-2x-2$ and $$nP(q^{2n})-P(q^2)=\frac{24KL}{\pi^2}\left\{1+kl+k'l'+\sqrt{kl}+\sqrt{k'l'}-\sqrt {klk'l'} \right\}$$ and we can verify that $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\sqrt{6}$$ for $n=19$. Using similar formulas given by Ramanujan we can verify the values for $n= 27, 35$, but for other values of $n$ there are no known formulas of similar kind to help us.

More generally if there is a formula of type $$nP(q^{2n})-P(q^2)=\frac {4KL}{\pi^2} \cdot A_n(k, l) $$ then $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\frac{G_n^{12}\cdot A_n(k, k') - \sqrt {n(G_n^{24}-1)}} {\sqrt{G_n^{24}-4} } $$ Ramanujan also gave formulas of type $$nP(-q^n) - P(-q) =\frac{4KL}{\pi^2}\cdot B_n(k,l)$$ for some odd positive integer values of $n$ and if such a formula is available then $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\frac{G_n^{12}\cdot B_n(k, k') } {2\sqrt {G_n^{24}-4} }$$

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There is an empirical approach which works nicely for $n=11,19,43,67,163$. If we evaluate the expression $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)$$ for $n=11,19$ then it is easy to spot that these values are $\sqrt{2},\sqrt{6}$ respectively.

Next for each of the values of $n$ mentioned in last paragraph one can prove that $G_n^8$ is a root of a polynomial $$x^3-a_nx^2-4$$ where $a_n$ is a positive integer. Further we can write $a_n=b_n^2\cdot c_n$ where $c_n$ is square free so that $$\sqrt{G_n^{24}-4}=\sqrt{x^3-4}=b_n x\sqrt{c_n}, x=G_n^8$$ and we can evaluate the expression $$\sqrt{\frac{c_n}{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)$$ numerically and confirm that it is an almost integer (depending on precision of numerical calculation) and let us say that the nearest integer is $d_n$. Then the expression in question is $d_n/\sqrt{c_n} $. Moreover for $n=43,67,163$ the expressions $P_n, Q_n$ are practically equal to $1$ and hence one just needs to evaluate the much simpler expression $$\sqrt{c_n} \left(\sqrt{n} - \frac{6}{\pi}\right)$$ and it turns out to be an almost integer. For completeness sake let us note that $c_n$ equals $2,6,15, 330,10005$ for $n=11,19,43,67,163$ respectively.

Having guessed the value of the expression under question empirically we can then try to prove it using approach discussed in first part of the answer.