Entry 37

There is an elegant continued fraction associated with the Ramanujan G and g functions$$G_n =\frac{2^{-1/4}\,\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\,\eta(2\tau)}$$ $$g_n =\frac{2^{-1/4}\,\eta\big(\tfrac{\tau}{2}\big)}{\eta(\tau)}$$where \(\tau=\sqrt{-n}\). Recall that \((G_n g_n)^8(G_n^8-g_n^8) = \tfrac{1}{4}\). So perhaps it is no surprise that they can be expressed by the octic continued fraction. Given the nome \(q=e^{\pi i \tau}\), then$$\beta(\tau)=\left(\frac{\sqrt{1+G_n^{-12}}\pm\sqrt{1-G_n^{-12}}}{2}\right)^{1/4} =\cfrac{\sqrt{2}\, q^{1/8}}{1 + \cfrac{q}{1 + q + \cfrac{q^2}{1 + q^2 + \cfrac{q^3}{1 + q^3 + \ddots}}}}$$where the \(\pm\) sign changes beyond a bound \(n\). Alternatively, \(\beta(\tau)=\left(-g_n^{12}+\sqrt{g_n^{24}+1}\right)^{1/4}\). For example, since \(G_{1/4} = \frac{(1+\sqrt{2})^{1/4}}{2^{3/16}}\) this implies the identity$$\left(\sqrt{\frac{1}{4}+\frac{2^{1/4}}{(1+\sqrt{2})^3}}-\sqrt{\frac{1}{4}-\frac{2^{1/4}}{(1+\sqrt{2})^3}}\right)^{1/4}=\sqrt{\frac{1}{1+\sqrt{2}}}$$ and is the value of the continued fraction when \(q=e^{\pi\, i\sqrt{-1/4}}\).

Entry 36

Given the nome \(p=e^{\pi i \tau}\) and \(\tau =\sqrt{-n}\), then the Ramanujan functions \(G_n\) and \(g_n\) are
$$\begin{aligned}G_n &= 2^{-1/4}p^{-1/24}\prod_{k=1,3,5,\dots}^\infty\big(1+p^k\big) \\ g_n &= 2^{-1/4}p^{-1/24}\prod_{k=1,3,5,\dots}^\infty\big(1-p^k\big)\end{aligned}$$ These are ubiquitous in Ramanujan's Notebooks. Well-known values are \(G_5 = \phi^{1/4}\) and \(G_{25} =\phi\), where \(\phi\) is the golden ratio. However, given the nome's square \(q=e^{2\pi i \tau}\), note also the three Weber modular functions,
$$\begin{aligned}\mathfrak{f}(\tau)&= q^{-1/48}\prod_{n=1}^\infty\big(1+q^{n-1/2}\big) \\ \mathfrak{f}_1(\tau)&= q^{-1/48}\prod_{n=1}^\infty\big(1-q^{n-1/2}\big) \\ \mathfrak{f}_2(\tau)&= \sqrt2\,q^{1/24}\prod_{n=1}^\infty\big(1+q^{n}\big) \end{aligned}$$ Notice the similarity of the definitions. Mathematica doesn't have built-in functions for these, but fortunately can be expressed by the more familiar Dedekind eta function \(\eta(\tau)\) with \(\tau =\sqrt{-n}\), $$G_n =2^{-1/4}\,\mathfrak{f}(\tau)=\frac{2^{-1/4}\,\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\,\eta(2\tau)}$$ $$g_n =2^{-1/4}\,\mathfrak{f}_1(\tau)= \frac{2^{-1/4}\,\eta\big(\tfrac{\tau}{2}\big)}{\eta(\tau)}$$The two functions \(G_n\) and \(g_n\) obey$$(G_n g_n)^8(G_n^8-g_n^8) = \tfrac{1}{4}$$which is consequence of Weber's$$\mathfrak{f}_1(\tau)^8+\mathfrak{f}_2(\tau)^8 = \mathfrak{f}(\tau)^8$$Ramanujan calculated many explicit values for \(G_n\) and \(g_n\), one of which is the remarkable \(G_{125}=3.6335\dots\)$$\frac{2\,G_{125}}{\varphi^{-5/4}}=-1+\frac{1}{\sqrt{5}}\left(1+2^{2/5}\varphi^{1/5}\left(\sqrt[5]{4-\sqrt{5}+(\varphi\sqrt{5})^{3/2}}+\sqrt[5]{4-\sqrt{5}-(\varphi\sqrt{5})^{3/2}}  \right)\right)^2$$ with the reciprocal golden ratio \(\varphi=\frac{-1+\sqrt{5}}{2}\approx0.61803\).

Entry 35

This continues Entry 34. Similar to its more famous cousin, Euler's prime-generating polynomial, the formula$$P(n)=6n^2-6n+31$$ yields consecutive prime values from \(n=0\to29\). Solving for \(P(n)=0\), one gets \(\tau=\frac{6+\sqrt{-708}}{12}=\frac{3+\sqrt{-177}}{6}\). Plugging this into $$\beta(\tau)=\Big (\left(\tfrac{\eta(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta(6\tau)}\right)^6-\left(\tfrac{\eta(\tau)\,\eta(6\tau)}{\eta(2\tau)\,\eta(3\tau)}\right)^6\Big)^2$$ with Dedekind eta function \(\eta(\tau)\), we find that it exactly yields an integer$$\beta\big(\tfrac{3+\sqrt{-177}}{6}\big)=-1060^2$$such that$$e^{2\pi/6\sqrt{177}} = 1060^2+9.999992\dots$$and the level-6 Ramanujan-Sato series$$\frac{1}{\pi}=\frac{1}{265\cdot1060}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}^3\,\frac{177\cdot7038k+89418}{(-1060^2)^k}$$with the binomial coefficient \(\binom{n}{k}\).

Entry 34

Here are some optimum prime-generating polynomials $$\begin{aligned}
P(n) &=n^2+n+41\\
P(n) &=2n^2+29\\
P(n) &=2n^2+2n+19\\
P(n) &=3n^2+3n+23\\
P(n) &=5n^2+5n+13\\
P(n) &=6n^2+6n+31\\
P(n) &=7n^2+7n+17
\end{aligned}$$The first one is the most famous, being Euler's. Expressed as the quadratic \(P(n) = an^2+bn+c\), its discriminant is \(d=b^2-4ac\). Using the values \(a,d\) of the above, one gets
$$\begin{aligned}
&e^{\pi/1\,\sqrt{163}} = 640320^3 +743.9999999999992\dots\\
&e^{\pi/2\,\sqrt{232}} = 396^4 -104.0000001\dots\\
&e^{\pi/2\,\sqrt{148}} = (84\sqrt{2})^4 +103.99997\dots\\
&e^{\pi/3\,\sqrt{267}} = 300^3 + 41.99997\dots\\
&e^{\pi/5\,\sqrt{5\times47}} = (18\sqrt{47})^2 + 15.991\dots\\
&e^{\pi/6\,\sqrt{6\times118}} = 1060^2 + 9.99992\dots\\
&e^{\pi/7\,\sqrt{7\times61}} = (39\sqrt{7})^2 + 9.995\dots
\end{aligned}$$These approximations (actually, the exact values of certain eta quotients) can be used as denominators in pi formulas known as Ramanujan-Sato series to be discussed in Entry 35.

 

Entry 33

In a similar manner to Entry 32, given the Chudnovsky algorithm$$\frac{1}{\pi} = \frac{12}{(640320)^{3/2}} \sum^\infty_{k=0} \frac{(6k)!}{(3k)!\,k!^3} \frac{3\cdot163\cdot1114806k + 13591409}{(-640320^3)^k}$$ we can use the cube root \(\sqrt[3]{-640320^3}=-640320\) as the median,$$\begin{aligned}\frac{1}{\pi}&=\frac{12}{(640320+12)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k(-3)^{k-3j}\tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j}\frac{163\cdot1114806k+(13591409-8\cdot181)}{(-640320-12)^k}\\ \frac{1}{\pi}&=\frac{12}{(640320-12)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\,3^{k-3j}\,\tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j}\,\frac{163\cdot1114806k+(13591409+8\cdot181) }{(-640320+12)^k}\end{aligned}$$ where \(\binom{n}{k}\) is the binomial coefficient. These are now level-9 Ramanujan-Sato series. 

Entry 32

It is well-known that$$\frac{1}{\pi} =\frac{2 \sqrt 2}{99^2} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot70\cdot13\,k+1103}{(396^4)^k}$$But it turns out we can also use the square root \(\sqrt{396^4}=\pm396^2\) as a median point for two other pi formulas. First express the above as$$\frac{1}{\pi} =\frac{192 \sqrt 2}{(396^2)^{3/2}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{2\cdot58\cdot15015k+72798}{(396^4)^k}$$then$$\begin{aligned}\frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2-16)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}\frac{58\cdot15015k+(72798+37)}{(-396^2+16)^k}\\ \frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+16)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}\frac{58\cdot15015k+(72798-37)}{(396^2+16)^k}\end{aligned}$$ where \(\binom{n}{k}\) is the binomial coefficient. Note that they have a beautifully symmetric form and how the same integers (which figure in Pell equations as discussed in Entry 1) appear in all three formulas. These two are level-8 Ramanujan-Sato series.