Ramanujan's take on Chudnovsky series for 1/π(PI): Part 2

In the previous post we have handled the evaluation of $P_n=P(-e^{-\pi\sqrt{n}}) $ for $n=11,27$. We will evaluate $P_n$ for $n=19,35$ in the current post and also discuss an empirical approach for $n=43,67,163$. Finally we will use the information in table given by Ramanujan to obtain certain series for $1/\pi$ (including the famous one by Chudnovsky brothers).

Evaluation of $P_n=P(-e^{-\pi\sqrt{n}}) =P(-q) $

To handle the case $n=19$ we start with a complicated modular equation of degree $19$ $$(\sqrt{kl} +\sqrt {k'l'} - 1)^5+ 112(\sqrt{kl}+\sqrt{k'l'}-1)^2\sqrt{klk'l'}\\-256(\sqrt{klk'l'}-\sqrt{kl} - \sqrt{k'l'})\sqrt{klk'l'} =0$$ Putting $l=k', k=l'$ we get $$(2\sqrt{kk'}-1)^5+112(2\sqrt{kk'}-1)^2kk'-256kk'(kk'-2\sqrt{kk'})=0$$ Putting $a=\sqrt{2}G_{19}^{-6}=2\sqrt{kk'}$ we get $$(a-1)^5+28(a-1)^2a^2-16a^4+64a^3=0$$ or $$a^5+7a^4+18a^3+18a^2+5a-1=0$$ Luckily $a=-1$ turns out to be a root and hence there is a factor $(a+1)$ involved here. We then have $$(a+1)(a^4+6a^3+12a^2+6a-1)=0$$ Since $a>0$ we have $$a^4+6a^3+12a^2+6a-1=0$$ Again we get a factor of $(a+1)$ and then $$(a+1)(a^3+5a^2+7a-1)=0$$ Since $a>0$ it turns out that $a=\sqrt{2}G_{19}^{-6}$ is a root of $$x^3+5x^2+7x-1\tag{1}$$ and the above polynomial is irreducible as neither $1$ nor $-1$ is a root.

Next we need the formula given by Ramanujan as $$nP(q^{2n})-P(q^2)=\frac{24KL}{\pi^2}\left\{1+kl+k'l'+\sqrt{kl}+\sqrt{k'l'}-\sqrt{klk'l'}\right\}\tag{2} $$ where $n=19$ and $k$ is of degree $19$ over $l$ so that nome $q$ corresponds to $l$ and $q^n$ to $k$. Putting $q=e^{-\pi/\sqrt{19}}$ so that $q^n=e^{-\pi\sqrt{19}}$ and $k=l', k'=l$ we get $$nP(q^{2n})-P(q^2)=6\left(\frac{2K}{\pi}\right)^2\frac{K'}{K}(1+kk'+2\sqrt{kk'})$$ or $$nP(q^{2n})-P(q^2)=6\sqrt{19}\left(\frac{2K}{\pi}\right)^2(1+(a^2/4)+a)=\frac{3\sqrt{19}}{2}\left(\frac{2K}{\pi}\right)^2(a+2)^2$$ Combining the above equation with the identity $$nP(q^{2n})+P(q^2)=\frac{6\sqrt{n}}{\pi}, q=e^{-\pi/\sqrt {n}}\tag{3} $$ we get $$2P(q^{2n})=\frac{6}{\pi\sqrt{n}}+\frac{3}{2\sqrt{19}}\left(\frac{2K}{\pi}\right)^2(a+2)^2$$ Switching notation such that $k, K$ correspond to nome $q=e^{-\pi\sqrt {19}}$ we get $$2P(q^2)-\frac{6}{\pi\sqrt {n}} =\frac{3}{2\sqrt{19}}\left(\frac{2K}{\pi}\right) ^2(a+2)^2$$ Using the identity $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)=\left(\frac{2K}{\pi}\right)^2\sqrt{1-G^{-24}}\tag{4}$$ we get $$\sqrt{n} P(-q) - \frac{6}{\pi}=\left(\frac{2K}{\pi}\right)^2\frac{3(a+2)^2-\sqrt {19(4-a^4)}}{2}$$ And then using identity $$Q(-q) = \left(\frac{2K}{\pi}\right)^4(1-4G^{-24})$$ we get $$\frac{1}{\sqrt{Q_n}}\left (\sqrt{n}P_n-\frac{6}{\pi}\right) =\frac{3(a+2)^2-\sqrt{19(4-a^4)}} {2\sqrt{1-a^4}}\tag{5}$$ To simplify the expression on right we evaluate some integral powers of $a$ \begin{align} a^3&=-5a^2-7a+1\notag\\ a^4&=-5a^3-7a^2+a=18a^2+36a-5\notag\\ \frac{1}{a}&=a^2+5a+7\notag\\ \frac{1}{a^2}&=a+5+\frac{7}{a}=7a^2+36a+54\notag \end{align} We can observe that $$\frac{1}{a^2}=a^2+6(a^2+6a+9)=a^2+6(a+3)^2$$ and thus on multiplying by $a^2$ we get $$1-a^4=6a^2(a+3)^2$$ ie $$\sqrt{1-a^4}=a(a+3)\sqrt{6}$$ Further $$4-a^4=9(1-4a-2a^2)$$ and hence we can write the expression on right of equation $(5)$ as $$\frac{3}{2}\cdot\frac{(a+2)^2-\sqrt{19(1-4a-2a^2)}}{a(a+3)\sqrt{6}}$$ and this equals the proposed value $\sqrt{6}$ if $$\sqrt{19(1-4a-2a^2)}=(a+2)^2-4a(a+3)=-3a^2-8a+4$$ Using numerical value of $a=0.16071$ we can confirm that the right hand side of above equation is positive and the above equation will hold true if the squares of both sides are equal ie if $$19(1-4a-2a^2)=(3a^2+8a-4)^2$$ or $$19(1-4a-2a^2)=9a^4+64a^2+16+48a^3-24a^2-64a$$ or $$3-12a-78a^2=9a^4+48a^3$$ or $$1-4a-26a^2=3a^4+16a^3$$ or $$1-4a-26a^2=3(18a^2+36a-5)+16(-5a^2-7a+1)$$ or $$1-4a-26a^2=1-4a-26a^2$$ which is true.

The case $n=35$ is a bit complicated as I haven't been able to find a suitable modular equation of degree $35$ which can be used to evaluate the class invariant $G_{35}$. Instead I use the table by Ramanujan to conclude that $G_{35}^8$ is a root of $$x^3-(60+28\sqrt {5})x^2-4\tag{6}$$ and its numerical value is around $122.61017$.

To simplify typing we use the symbol $a=\sqrt {5}$ in handling the case $n=35$. Next we can note that $b=G_{35}^4$ is a root of $$x^6-(60+28a)x^4-4$$ and the above polynomial can be factored as $$(x^3-(6+2a)x^2-2(1+a)x-2)(x^3+(6+2a)x^2-2(1+a)x+2)$$ in polynomial ring $\mathbb{Q} (a) [x] $. The above factorization was done using Magma calculator online and one can verify it manually. The first factor has a positive root and thus $b=G_{35}^4$ is a root of $$x^3-(6+2a)x^2-2(1+a)x-2\tag{7}$$ Replacing $x$ with $x^2$ doesn't lead to any further factorization and thus $G_{35}^2$ is of degree $6$ over $\mathbb{Q} (a) $.

Luckily we can note that $2b$ is a root of $$x^3-(12+4a)x^2-(8+8a)x-16$$ and replacing $x$ with $x^2$ gives us the factorization (again via Magma) $$(x^3-4x^2+2(1-a)x-4)(x^3+4x^2+2(1-a)x+4)$$ with first factor having a positive root and thus $\sqrt {2}G_{35}^2$ is a root of $$x^3-4x^2+2(1-a)x-4$$ It follows that $c=G_{35}^2$ is a root of $$x^3-2\sqrt{2}x^2+(1-a)x-\sqrt{2}$$ and hence $$\sqrt {2}=\frac{c(c^2+1-a)}{2c^2+1}=\frac{c(b+1-a)}{2b+1}$$ as $b=c^2$. Therefore $$\frac{\sqrt{2}} {c} =\sqrt{\frac{2} {b}}=\frac{b+1-a}{2b+1}\tag{8}$$ is a rational function of $b$ with coefficients in $\mathbb{Q} (a) $ and this fact will be used later.

We also need to evaluate the expression $1/(2b+1)$ as a polynomial in $b$. While one can use Magma for this, the verification by hand is bit lengthy and hence we use algebra directly. We note that $(2b+1)$ is a root of $$(x-1)^3-(12+4a)(x-1)^2-(8+8a)(x-1)-16 $$ or $$x^3-(15+4a)x^2+19x-(21-4a)$$ and hence $$\frac{21-4a}{2b+1}=(2b+1)^2-(15+4a)(2b+1)+19=4b^2-(26+8a)b+5-4a$$ and noting that $$(21-4a)(21+4a)=19^2 $$ we get $$\frac{1}{2b+1}=\frac{1}{19^2}((84+16a)b^2-(706+272a)b+25-64a)$$ Next we evaluate the expression $\sqrt{2/b}=(b+1-a)/(2b+1)$ and this turns out to be simpler in form than $1/(2b+1)$ and we have $$\sqrt{\frac{2}{b}}=\frac{b+1-a}{2b+1}=\frac{1}{19}(-2(1+2a)b^2+(53+30a)b+3(9-a))\tag{9}$$
We now use the formula given by Ramanujan $$nP(q^{2n})-P(q^2)\\=\frac{4KL}{\pi^2}\left\{2(\sqrt{kl}+\sqrt{k'l'}-\sqrt {klk'l'}) +(4klk'l')^{-1/6}(1-\sqrt{kl}-\sqrt{k'l'})^3\right \}$$ for $n=35$. Here we assume $k$ to be of degree $35$ over $l$ with $q$ being nome corresponding to $l$ and $q^n$ the nome corresponding to $k$. Putting $q=e^{-\pi/\sqrt {35}}$ so that $q^n=e^{-\pi\sqrt{35}}$ and $k=l', l=k'$ and $L/K=K'/K=\sqrt {35}$ we get $$nP(q^{2n})-P(q^2)=\sqrt {35}\left (\frac{2K}{\pi}\right) ^2\left(2(2\sqrt {kk'} - kk') +(4k^2k'^2)^{-1/6}(1-2\sqrt{kk'})^{3}\right)$$ Using $2kk'=G_{35}^{-12}$ we get $$nP(q^{2n})-P(q^2)=\sqrt {35}\left(\frac{2K}{\pi}\right)^2\left(2(\sqrt{2}G_{35}^{-6}-G_{35}^{-12}/2)+G_{35}^4(1-\sqrt{2}G_{35}^{-6})^3\right)$$ We now have $$\sqrt{2}G_{35}^{-6}=\frac{\sqrt {2}}{G_{35}^2\cdot G_{35}^4}=\frac{1}{b}\sqrt{\frac{2}{b}}$$ and then $$nP(q^{2n})-P(q^2)=\sqrt {35}\left(\frac{2K}{\pi}\right)^2\left\{2\left(\frac{1}{b}\sqrt{\frac{2}{b}}-\frac{1}{2b^3}\right)+b\left(1- \frac{1}{b}\sqrt{\frac{2}{b}} \right)^3\right\}$$ Combining this with identity $$nP(q^{2n})+P(q^2)=\frac{6\sqrt {n}} {\pi}, q=e^{-\pi/\sqrt{n}} $$ we get $$2P(q^{2n})=\frac{6}{\pi\sqrt{n}}+\frac{1}{\sqrt{35}}\left(\frac{2K}{\pi}\right)^2\left\{2\left(\frac{1}{b}\sqrt{\frac{2}{b}}-\frac{1}{2b^3}\right)+b\left(1-\frac{1}{b}\sqrt{\frac{2}{b}}\right)^3\right\}$$ Changing the notation a bit so that $q=e^{-\pi\sqrt{35}}$ corresponds to $k$ we have $$2P(q^2)=\frac{6}{\pi\sqrt{n}}+\frac{1}{\sqrt{35}}\left(\frac{2K}{\pi}\right)^2\left\{2\left(\frac{1}{b}\sqrt{\frac{2}{b}}-\frac{1}{2b^3}\right)+b\left(1-\frac{1}{b}\sqrt{\frac{2}{b}}\right)^3\right\}$$ Using the identity $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)=\left(\frac{2K}{\pi}\right)^2\sqrt{1-G^{-24}}$$ we get $$\sqrt{n} P(-q)-\frac{6}{\pi}=\frac{1}{b^3}\left(\frac{2K}{\pi}\right)^2\left\{2b^2\sqrt{\frac{2}{b}}-1+b\left(b-\sqrt{\frac{2}{b}}\right)^3-\sqrt{35(b^6-1)}\right\}$$ and finally using $$Q(-q) =\left(\frac{2K}{\pi}\right)^4(1-4G^{-24})$$ we get $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n}P_n-\frac{6}{\pi}\right)=\frac{1}{\sqrt{b^6-4}}\left\{2b^2\sqrt{\frac{2}{b}}-1+b\left(b-\sqrt{\frac{2}{b}}\right)^3-\sqrt{35(b^6-1)}\right\}$$ We have to show that this expression equals $$(2+\sqrt{5})\sqrt{\frac{2}{\sqrt{5}}}=(2+a) \sqrt{\frac{2}{a}}$$ Using equation $(6)$ we have $$b^6-4=(60+28a)b^4$$ and $$\sqrt{(60+28a)(2/a)}=\sqrt{56+24a}=6+2a$$ and thus we have to establish that $$2b^2\sqrt{\frac{2}{b}}-1+b\left(b-\sqrt{\frac{2}{b}}\right)^3-\sqrt{35(b^6-1)}=(2+a) (6+2a)b^2=(22+10a)b^2$$ or $$2b^2\sqrt{\frac{2}{b}}+b\left(b-\sqrt{\frac{2}{b}}\right)^3-(22+10a)b^2-1=\sqrt{35(b^6-1)}$$ We have $$b-\sqrt{\frac{2}{b}}=\frac{1}{19}(2(1+2a)b^2-2(17+15a)b-3(9-a))$$ and $$2b^2\sqrt{\frac{2} {b}} =\frac{1}{19}(2(9-a)b^2+4(9-a)b+4(1+2a))$$ Next we have $$\left(b-\sqrt{\frac{2}{b}} \right) ^2=\frac{1}{19}(4(9-a)b^2-8(10+a)b-2(15+11a))$$ and $$\left(b-\sqrt{\frac{2}{b}}\right) ^3=\frac{4}{19}((15+11a)b^2+2(15+11a)b+2(8-3a))$$ and $$b\left(b-\sqrt{\frac{2}{b}}\right) ^3=\frac{8}{19}((115+59a)b^2+(78+23a)b+(15+11a))$$ Using these expressions above we get $$2b^2\sqrt{\frac{2}{b}}+b\left(b-\sqrt{\frac{2}{b}}\right)^3-(22+10a)b^2-1\\=\frac{1}{19}((520+280a)b^2+(660+180a)b+(105+96a))\tag{10}$$ The expression on right is a positive real number (because $a, b$ are positive as well) and hence our job is done if we show that square of this number equals $$35(b^6-1)=35((22+10a)b^4+3)\\=35((7120+3184a)b^2 + (4280+1912a)b + 1283+576a) $$ This is difficult to verify by hand. Luckily we can note that $$\sqrt{35(b^6-1)}=\sqrt{5(b^3-1)\cdot 7(b^3+1)}$$ and $$b^3-1=(6+2a)b^2+2(1+a)b+1=((1+a)b+1)^2$$ and hence we have $$\sqrt{35(b^6-1)}=a((1+a)b+1)\sqrt{7(b^3+1)}$$ We can thus try to divide the expression on right hand side of equation $(10) $ by $a((1+a)b+1)$ and then show that square of resulting expression is $7(b^3+1)$. Our hope is that the calculations involved will be manageable by hand.

To begin with we first need to deal with reciprocal of $(1+a)b+1$. Let us observe that $(1+a)b$ is root of $$x^3-(6+2a)(1+a)x-2(1+a)^3x-2(1+a) ^3$$ or $$x^3-8(2+a)x^2-16(2+a)x-16(2+a) $$ and then $(1+a)b+1$ is root of $$(x-1)^3-(16+8a)(x-1)^2-(32+16a)(x-1)-(32+16a)$$ or $$x^3-(19+8a)x^2+3x-(17+8a)$$ We then have $$\frac{17+8a}{(1+a)b+1}=((1+a)b+1)^2-(19+8a)((1+a)b+1)+3\\=(6+2a)b^2-(57+25a)b-(15+8a)$$ We note that $$(17+8a)(8a-17)=31$$ and hence $$\frac{1}{(1+a)b+1}=\frac{1}{31}((-22+14a)b^2-31(1+a)b-(65-16a)) $$ We next have $$(520+280a)b^2+(660+180a)b+105+96a\\ =20(26+14a)b^2+60(11+3a)b+105+96a$$ which is same as $$20(11+3a)b[(1+a)b+1]+40(11+3a)b+105+96a$$ and this can be written as $$20(11+3a)b[(1+a)b+1]+40(1+2a)[(1+a)b+1]+65+16a$$ It now follows that $$\frac{(520+280a)b^2+60(11+3a)b+105+96a} {(1+a)b+1}\\=20(11+3a)b+40(1+2a)+\frac{65+16a} {(1+a)b+1}$$ Noting that $$(65+16a)(65-16a)=31\cdot 95$$ we get $$\frac{65+16a}{(1+a)b+1}=(18a-10)b^2-(145+81a)b-95$$ and then $$\frac{(520+280a)b^2+60(11+3a)b+105+96a} {(1+a)b+1} \\=(18a-10)b^2+(75-21a)b+80a-55$$ Dividing by $a$ we get $$\frac{(520+280a)b^2+60(11+3a)b+105+96a} {a((1+a)b+1)} \\ =2(9-a)b^2-3(7-5a)b+(80-11a)$$ We next compute the square of this expression and show that it equals $19^2\cdot 7(b^3+1)$ and thereby our job is done.

We have $$[2(9-a)b^2-3(7-5a)b+(80-11a)] ^2\\=4(9-a)^2b^4+9(7-5a)^2b^2+(80-11a)^2-12(9-a)(7-5a)b^3\\-6(7-5a)(80-11a)b+4(9-a)(80-11a)$$ The above expression equals $$8(43-9a)b^4+18(87-35a)b^2+(7005-1760a)\\-48(22-13a)b^3-6(835-477a)b+4(775-179a)b^2\tag{11} $$ We note that $$b^3=2(3+a)b^2+2(1+a)b+2$$ and \begin{align} b^4&=2(3+a)b^3+2(1+a)b^2+2b\notag\\ &=2(3+a)[2(3+a)b^2+2(1+a)b+2]+2(1+a)b^2+2b\notag\\ &=8(7+3a)b^2+16(2+a)b+4(3+a)+2(1+a)b^2+2b\notag\\ &=2(29+13a)b^2+2(17+8a)b+4(3+a) \notag \end{align} Using these values of $b^3,b^4$ we find that expression in $(11) $ equals $$16(662+298a)b^2+16(371+191a)b+32(84+16a)\\-96(1-17a)b^2-96(-43+9a)b-96(22-13a)\\+(4666-1346a)b^2-6(835-477a)b+7005-1760a)$$ or $$16(662+298a)b^2+16(371+191a)b+32(84+16a)\\-96(1-17a)b^2-96(-43+9a)b-96(22-13a)\\+(4666-1346a)b^2-6(835-477a)b+7005-1760a)$$ or $$(15162+5054a)b^2+(5054+5054a)b+(7581)$$ The above expression can be written as $$2527[(6+2a)b^2+2(1+a)b+3]=19^2\cdot 7(b^3+1)$$ and we are done now.

The Magma code below creates the field extensions $K=\mathbb{Q} (a), L=K(b) $ and then one can evaluate expressions in field $L$ as polynomials in $b$ with coefficients in $K$. The code can be run using online Magma calculator.

R<x>:= PolynomialRing(Integers()) ;
f := x^2-5;
K<a> := NumberField(f) ;
T<y> := PolynomialRing(K) ;
Factorization(y^6-(60+28*a)*y^4-4);
g := y^3-(6+2*a)*y^2-2*(1+a)*y-2;
Factorization(y^6-(12+4*a)*y^4-(8+8*a)*y^2-16);
L<b> := ext<K|g>;
Sqrt(2/b);
(b+1-a)/(2*b+1);
(2*b^2*Sqrt(2/b)+b*(b-Sqrt(2/b))^3-(22+10*a)*b^2-1)/(a*((1+a)*b+1));
Sqrt(7*(b^3+1));

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 $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\frac{G_n^{12}\cdot B_n(k, k') } {2\sqrt {G_n^{24}-4} }$$

An Empirical Approach for Evaluation of $P_n=P(-e^{-\pi\sqrt{n}})$

There is also an empirical approach which works nicely for $n=11,19,43,67,163$. For each of these values of $n$ one can show that $G_n^8$ is a root of the equation $$x^3-u_nx^2-4$$ where $u_n$ is a positive integer. More specifically we have $$u_n=8,24,240,1320,160080$$ for the values of $n$ being considered respectively. And then we get $$\sqrt {G_n^{24}-4}=\sqrt {u_n} G_n^8$$ And one can check that $$d_n=\sqrt{\frac{u_n}{Q_n}} \left(\sqrt{n} P_n-\frac{6}{\pi}\right)$$ is a positive integer as well. Thus one can try to evaluate the above expression numerically and take the nearest integer to this numerical value as $d_n$ and make an empirical claim that $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\frac{d_n}{\sqrt{u_n}}$$ Further for $n=43,67,163$ the expressions $P_n, Q_n$ are practically equal to $1$ and it is sufficient to evaluate just $$\sqrt{u_n} \left(\sqrt{n} - \frac{6}{\pi}\right)$$ and take $d_n$ to be nearest integer. The empirical claim however needs to be verified analytically as we have done above.

Let us further note that $G_n^4$ is a root of $$x^6-u_nx^4-4$$ and this can be factored into two cubics as $$(x^3+px^2+qx-2)(x^3-px^2+qx+2)$$ and using a bit of algebra one can evaluate $p, q$. More spefically if $v_n$ is a root of $$x^4-32x-16u_n$$ then $G_n^4$ is a root of $$x^3-(v_n^2/4)x^2+v_nx-2$$ For the values of $n$ being considered here $v_n$ turns out to be an even integer and then $$\sqrt{G_n^{12}-1} =\pm((v_n/2)G_n^4-1)$$ where the $\pm$ sign is same as that of $v_n$. For completeness sake let us note that $$v_n=4,-4, 8,-12,-40$$ for the above mentioned values of $n$ respectively.

For the specific case of $n=35$ both $u_n, v_n$ turn out to be algebraic integers from $\mathbb{Q} (\sqrt{5})$ with $$u_n=60+28\sqrt{5},v_n=-2(1+\sqrt{5})$$ Also when we evaluate the expression $$\sqrt{60+28\sqrt{5}}\left(\sqrt{35}-\frac{6}{\pi}\right)$$ numerically we get $44.3606214$ and the first four decimal digits match those of $10\sqrt{5}$. Subtracting $10\sqrt{5}$ from the above result we get $21.9999416$ and thus the expected value of the expression $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)$$ is $$\frac{22+10\sqrt{5}}{\sqrt{60+28\sqrt{5}}}=(2+\sqrt{5})\sqrt{\frac{2}{\sqrt{5}}}$$ and we can verify this analytically as mentioned earlier.

While evaluating $P_n$ we have to deal with expressions like $$\sqrt{b^6-4},\sqrt{n(b^6-1)}$$ where $b=G_n^4$. We have mentioned earlier that $$\sqrt{b^6-4}=\sqrt{u_n}b^2,\sqrt{b^3-1}=\operatorname{sgn}(v_n)((v_n/2)b-1)$$ and it turns out that the expression $\sqrt{n(b^3+1)}$ also lies in $\mathbb{Q} (b) $ so that $\sqrt{n(b^6-1)}\in\mathbb{Q}(b)$ and this strange coincidence helps us verify the exact radical expression for $P_n$.

$P(-q)$ and Chudnovsky series for $1/\pi$

The values of $P_n=P(-q),Q_n=Q(-q),R_n=R(-q),q=e^{-\pi\sqrt{n}}$ in the table given by Ramanujan indicate that Ramanujan had managed to discover some more series for $1/\pi$ (apart from those listed in his famous paper Modular equations and approximations to $\pi$), although he never wrote them explicitly. And these series famously include the one given by Chudnovsky brothers.

Now I show how the information in the page from Ramanujan's lost notebook can be used to obtain series for $1/\pi$. It is also important to understand that Ramanujan presented the values in table in exactly the form needed to get these series. In particular the coefficient of $Q_n^3$ has a factor $n$ and Ramanujan specifically mentions that for $n=163$ we have $$53360^3+1=3^3\cdot 7^2\cdot 11^2\cdot 19^2\cdot 127^2\cdot 163$$ The factorization for coefficient of $Q_n^3$ is needed to get the desired series as will be shown below.

Let us start with hypergeometric identity $$\left(\frac{2K}{\pi}\right)^2=\frac{1}{\sqrt{1-4G^{-24}}}{}_3F_2\left(\frac{1}{6},\frac{5}{6},\frac{1}{2};1,1;-\frac{27G^{-24}}{(1-4G^{-24})^3}\right)\tag{12}$$ Also let us use the simplified notation $$f(x) ={}_3F_2\left(\frac{1}{6},\frac{5}{6},\frac{1}{2};1,1;x\right),g(x) =xf'(x) \tag{13}$$ in what follows.

Using derivatives of elliptic integrals we can prove that $$P(-q) = \left(\frac{2K}{\pi}\right)^2(1-2k^2)+3kk'^2\frac{d}{dk} \left(\frac{2K}{\pi}\right)^2\tag{14}$$ Applying the above formula on equation $(12)$ and in the process doing reasonable amount of symbol shunting we arrive at the following beautiful result $$P(-q) =\frac{(1-2k^2)(1+8G^{-24})}{(1-4G^{-24})^{3/2}}\left(f(v(k))+6g(v(k))\right),v(k)=-\frac{27G^{-24}}{(1-4G^{-24})^3}\tag{15}$$ Next let $n>3$ be an odd positive integer and we use $q=e^{-\pi\sqrt{n}} $ so that $$P_n=P(-q), Q_n=Q(-q), R_n=R(-q), G_n=G$$ If we carefully observe the formula $(15)$ we notice at once that the first factor on right is $$\frac{(1-2k^2)(1+8G_n^{-24})}{(1-4G_n^{-24})^{3/2}}=\sqrt {\frac{R_n^2} {Q_n^3}}$$ and the values of the right hand side are given using the relationship between $Q_n, R_n$ provided by Ramanujan. It should be noted that the factorization of coefficient of $Q_n^3$ (given in relation between $Q_n, R_n$) helps in finding the square root on right side of the above equation.

Let us now set $$p_n=\frac{1}{\sqrt{Q_n}}\left(\sqrt{n}P_n-\frac{6}{\pi}\right)$$ so that $$P_n=\frac{6}{\pi\sqrt{n}}+\frac{p_n\sqrt{Q_n}}{\sqrt{n}}=\frac{6}{\pi\sqrt{n}}+\frac{p_n}{\sqrt{n}}\left(\frac{2K}{\pi}\right)^2\sqrt{1-4G_n^{-24}}$$ and using equation $(14), (15)$ we get $$P_n= \frac{6}{\pi\sqrt{n}}+\frac{p_n}{\sqrt{n}}\cdot f(v(k)) \tag{16}$$ Comparing equations $(15),(16)$ and using the values $p_n$ and the relations between $Q_n, R_n$ we get desired series for $1/\pi$ as $$\frac{6}{\pi}=\left(\sqrt{\frac{nR_n^2}{Q_n^3}}-p_n\right)f(v(k))+6\sqrt{\frac{nR_n^2}{Q_n^3}}g(v(k))\tag{17}$$ For $n=11,19,43,67,163$ the relation between $Q_n, R_n$ and value $v(k) $ is easily found. One can prove with some effort that for each of these values of $n$ the value $x_n=G_n^8$ is a root of $$x^3-u_nx^2-4=0$$ where $u_n$ is a positive integer and $y_n=x_n^3=G_n^{24}$ is a root of $$y^3 - (u_n^3 + 12)y^2 + 48y - 64=0$$ The values of $u_n$ are $$u_n=8,24,240,1320,160080$$ for the above values of $n$ respectively. Let us now note that $$v(k) =-\frac{27G_n^{-24}}{(1-4G_n^{-24})^3}=-\frac{27y_n^2}{(y_n-4)^3}=-\frac{27}{u_n^3}$$ And further we have \begin{align} \frac{R_n^2} {Q_n^3}& =\frac{(1-2k^2)^2(1+8G_n^{-24})^2}{(1-4G_n^{-24})^{3}}\notag\\ &=\frac{(1-G_n^{-24})(1+8G_n^{-24})^2} {(1-4G_n^{-24}) ^3} \notag\\ &=\frac{y_n^3+15y_n^2+48y_n-64} {(y_n-4)^3 } \notag\\ &=1+\frac{27}{u_n^3}\notag \end{align} so that using the values of $u_n$ the relations between $Q_n, R_n$ are available.

Using these values the series $(17)$ becomes $$\frac{6u_n\sqrt{u_n}}{\pi}\\=\left(\sqrt{n(u_n^3+27)}-p_nu_n\sqrt{u_n}\right)f\left(-\frac{27}{u_n^3}\right)+6\sqrt{n(u_n^3+27)}\cdot g \left(-\frac{27}{u_n^3}\right) \tag{18}$$ It should be noted that the expressions $$\sqrt{n(u_n^3+27)},p_n\sqrt{u_n}$$ are positive integers for these particular values of $n$ (we have discussed the empirical approach to evaluate the value of $p_n$ using the fact that $p_n\sqrt{u_n} $ is a positive integer earlier). For $n=35$ we have $u_n=60+28\sqrt{5}$ and these expressions are algebraic integers from field $\mathbb{Q} (\sqrt{5})$. For $n=27$ we have $u_n^3=192000$ and $$\sqrt {n(u_n^3+27)},p_n\sqrt{u_n^3}$$ are positive integers.

The above formula holds for all positive integers $n>3$ with $u_n=(G_n^{24}-4)/G_n^{16}$ and in practice one uses odd values of $n$ because the evaluation and form of $G_n$ is simpler for odd $n$. Thus for example we have $u_n=15/4,p_n=\sqrt{3/5}$ for $n=7$. It turns out that for $n=7,11,19,27,43,67,163$ the series for $1/\pi$ involve only one quadratic irrationality and other numbers appearing in the series are rational.

We exhibit two series for $1/\pi$ based on above approach for $n=11,163$. For $n=11$ we have $u_n=8,p_n=\sqrt{2}$ and $$ \sqrt{n(u_n^3+27)}-p_nu_n\sqrt{u_n}=\sqrt{11\cdot 539}-32=45$$ and $$ \sqrt{n(u_n^3+27)}=77$$ and using equation $(18)$ we get the desired series as $$\frac{96\sqrt{2}}{\pi}=45f(-27/512)+6\cdot 77g(-27/512) $$ or $$\frac{32\sqrt{2}}{\pi}=15f(-27/512)+154g(-27/512)$$ or $$\frac{32\sqrt{2}}{\pi}=\sum_{j\geq 0}(-1)^j\frac{(1/6)_j(5/6)_j(1/2)_j} {(j!) ^3}(15+154j)\left(\frac{3} {8}\right)^{3j}$$ For $n=163$ we have $$u_n=160080,p_n=362\sqrt{\frac{3}{3335}}$$ so that $$\sqrt{n(u_n^3+27)}-p_nu_n\sqrt{u_n}\\=\sqrt{163\cdot 27(53360^3+1)}-160080\cdot 362\sqrt{\frac{3}{3335}}\sqrt{160080}\tag{19}$$ Now we use the factorization $$53360^3+1=3^3\cdot 7^2\cdot 11^2\cdot 19^2\cdot 127^2\cdot 163$$ given by Ramanujan to simplify the expression in $(19)$ as $$27\cdot 163\cdot 7\cdot 11\cdot 19\cdot 127-160080\cdot 362\cdot 12=122322681$$ and we have $$\sqrt{n(u_n^3+27)} =817710201$$ and using equation $(18)$ we get $$\frac{6\cdot 160080\sqrt {160080}}{\pi} =122322681f(-1/53360^3)+817710201\cdot 6g(-1/53360^3)$$ or $$\frac{106720\sqrt{160080}}{\pi}=\sum_{j\geq 0}(-1)^j\frac{(1/6)_j(5/6)_j(1/2)_j}{(j!)^3}\frac{13591409+545140134j} {53360^{3j}}$$ which is the famous series by Chudnovsky brothers.

Note: The above presentation is based on a thread on MathOverflow I posted sometime ago.

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