Entry 31

This was inspired by Entry 30. My version is a solution to $$\big(\tfrac{1}{2}\sin x\cos z\big)^{1/4}+\big(\tfrac{1}{2}\cos x\sin z\big)^{1/4} =\big(\sin 2y)^{1/12}$$ can be given by \(x=4\arctan u_1\) and \(z=4\arctan u_2\) where the \(u_i\) are appropriate roots of the quartic$$a^2u^4+\tfrac{1}{a}(1-3a^2)u^3+3(1-a^2)u^2+a(3-a^2)u-1=0$$and \(a=\tan(y/4)\) for real \(0<y\leq1\). For example, let \(y=1\) so \(a=\tan(1/4)\approx 0.2553\), then \(u_1,u_2\) are the two real roots of the quartic.

Alternatively, let \(u_1,u_2\) be the same roots and define the Pythagorean triples$$\begin{aligned}p_1,\, p_2,\, p_3 &= 2u_1,\; u_1^2-1,\; u_1^2+1\\ q_1,\, q_2,\, q_3 &= 2u_2,\; u_2^2-1,\; u_2^2+1\\r_1,\, r_2,\, r_3 &= 2a,\; a^2-1,\; a^2+1\end{aligned}$$Then for some constant \(0<a<(-1+\sqrt{2})\), or \(1<a<(1+\sqrt{2})\), we have$$\left(\frac{p_1p_2(q_1^2-q_2^2)}{p_3^2q_3^2} \right)^{1/4}+\left(\frac{q_1q_2(p_1^2-p_2^2)}{p_3^2q_3^2} \right)^{1/4}=\left(\frac{4r_1r_2(r_1^2-r_2^2)}{r_3^4} \right)^{1/12}$$The equality holds, but I don't know why it works.

Entry 30

There's this unusual trigonometric relation. If,$$\sin(x+y) = 2\sin\big(\tfrac{1}{2}(x-y)\big)$$$$\sin(y+z) = 2\sin\big(\tfrac{1}{2}(y-z)\big)$$then$$\big(\tfrac{1}{2}\sin x\cos z\big)^{1/4}+\big(\tfrac{1}{2}\cos x\sin z\big)^{1/4} =\big(\sin 2y)^{1/12}$$The example by Ramanujan was,$$\begin{aligned}x &= \frac{\pi-\arcsin(p)}{2}=1.094\dots,\quad p=(4+\sqrt{15})^2\,\phi^{-9}\\ y &=\frac{\arcsin\big(\phi^{-3}\big)}{2}=0.119\dots\\z &=\frac{\arcsin(q)}{2}=0.0001\dots,\quad q=(4-\sqrt{15})^2\,\phi^{-9}\end{aligned}$$with golden ratio \(\phi=\frac{1+\sqrt{5}}{2}\). This implies the remarkable identity in radicals$$\Big(\tfrac{(\sqrt{1+p}\,+\sqrt{1-p})\,(\sqrt{1+q}\,+\sqrt{1-q})}{2^3}\Big)^{1/4}+\Big(\tfrac{(\sqrt{1+p}\,-\sqrt{1-p})\,(\sqrt{1+q}\,-\sqrt{1-q})}{2^3}\Big)^{1/4}=\phi^{-1/4}$$ with \(p=(4+\sqrt{15})^2\,\phi^{-9}\), and \(q=(4-\sqrt{15})^2\,\phi^{-9}\) as defined above. This was Question 359 of JIMS 4, p.78. (See The Problems Submitted by Ramanujan to the Journal of the Indian Mathematical Society, p. 9, by Bruce Berndt, et al.)

Entry 29

The first Watson triple integral was discussed in Enry 26. It turns out all three integrals can be expressed simply by the Dedekind eta function \(\eta(\tau)\)$$\begin{aligned}I_1 &= \frac{1}{\pi^3}\int_0^\pi \int_0^\pi \int_0^\pi \frac{dx\, dy\, dz}{1-\cos x\cos y\cos z}=4\,\eta^4(i)\\I_2 &= \frac{1}{\pi^3}\int_0^\pi \int_0^\pi \int_0^\pi \frac{dx\, dy\, dz}{3-\cos x\cos y-\cos x\cos z-\cos y\cos z}=4^{1/3}\sqrt{3}\,\eta^4(\sqrt{-3})\\I_3 &= \frac{1}{\pi^3}\int_0^\pi \int_0^\pi \int_0^\pi \frac{dx\, dy\, dz}{1-\cos x-\cos y-\cos z}=2\sqrt{6}\,(1+\sqrt{2})^{1/3}\eta^4(\sqrt{-6})\end{aligned}$$In terms of the gamma function$$\begin{aligned}I_1&= \frac{\Gamma^4(\frac{1}{4})}{4\pi^3}\\I_2&= \frac{3\Gamma^6(\frac{1}{3})}{2^{14/3}\pi^4}\\I_3&= \frac{\Gamma(\frac{1}{24})\Gamma(\frac{5}{24})\Gamma(\frac{7}{24})\Gamma(\frac{11}{24})}{16\sqrt{6}\,\pi^3} \end{aligned}$$. There are infinitely many Ramanujan-type formulas for the \(I_i\) a few of which are $$\begin{aligned}\frac{(1+\sqrt{2})\Gamma\big(\tfrac{3}{8}\big)^4}{(2\pi)^{5/2}}\sqrt{I_1} &= \sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(-2^6)^n}\\
\frac{1}{\sqrt{2}}\,I_1 &=\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(-2^9)^n}\\
\frac{4}{\sqrt{3}}\,I_2 &=\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^8)^n}\\
\frac{1+\sqrt{2}+\sqrt{6}}{\sqrt{6}}\,I_3 &=\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{\big(2^3(1+\sqrt{2}+\sqrt{6})^3(1+\sqrt{2})\big)^n}\end{aligned}$$ There is one other formula that belongs to the family with denominators as powers of \(2\)$$16\,\eta(\sqrt{-7})^4 =\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^{12})^n}$$but I don't know if this has an equivalent and analogous integral.

Entry 28

Ramanujan gave the infinite series,$$1 - \left(\frac{1}{2}\right)^3 + \left(\frac{1\times3}{2\times4}\right)^3 - \left(\frac{1\times3\times5}{2\times4\times6}\right)^3 + \cdots =\left(\frac{\Gamma\big(\tfrac{9}{8}\big)}{\Gamma\big(\tfrac{5}{4}\big)\Gamma\big(\tfrac{7}{8}\big)}\right)^2=\frac{8\,\big[K\big(k_2\big)\big]^2}{(1+\sqrt{2})\,\pi^2}$$$$1 - 5\left(\frac{1}{2}\right)^3 + 9\left(\frac{1\times3}{2\times4}\right)^3 - 13\left(\frac{1\times3\times5}{2\times4\times6}\right)^3 + \cdots = \frac{2}{\pi}$$$$1 + 9\left(\frac{1}{4}\right)^4 + 17\left(\frac{1\times5}{4\times8}\right)^4 + 25\left(\frac{1\times5\times9}{4\times8\times12}\right)^4 + \cdots =\frac{2^{3/2}}{\pi^{1/2}\,\Gamma^2\left(\frac{3}{4}\right)}=\frac{2^{5/2}}{\pi} \eta^2(i)$$with elliptic integral singular value \(K(k_2)\) and Dedekind eta function \(\eta(\tau)\). For the first and second, their apparent simplicity belies its deep connection to modular forms and there are in fact an infinite number of such formulas. The three can be succinctly expressed as$$\begin{aligned}S_1&=\sum_{n=0}^\infty\, (-1)^n \left(\frac{\Gamma\big(n+\tfrac{1}{2}\big)}{n!\;\Gamma\big(\tfrac{1}{2}\big)}\right)^3\\S_2&=\sum_{n=0}^\infty\, (-1)^n\,(4n+1) \left(\frac{\Gamma\big(n+\tfrac{1}{2}\big)}{n!\;\Gamma\big(\tfrac{1}{2}\big)}\right)^3\\S_3&=\sum_{k=0}^\infty (8k+1)\left(\frac{\Gamma\big(k+\tfrac{1}{4}\big)}{k!\;\Gamma\big(\tfrac{1}{4}\big)}\right)^4\end{aligned}$$However, since$$\frac{\Gamma\big(n+\tfrac{1}{2}\big)}{n!\;\Gamma\big(\tfrac{1}{2}\big)}=\frac{(2n)!}{2^{2n}\,n!^2}=\frac{(2n-1)!!}{(2n)!!}$$then \(S_2\) is equivalently,$$\frac{2}{\pi}=\sum_{n=0}^\infty\,(-1)^n \left(\frac{(2n)!}{n!^2}\right)^3 \frac{4n+1}{2^{6n}}$$Another that belongs to the same family is$$\frac{16}{\pi}=\sum_{n=0}^\infty \left(\frac{(2n)!}{n!^2}\right)^3 \frac{42n+5}{2^{12n}}$$and so on. For the third, note that$$\frac{\Gamma\big(n+\tfrac{1}{4}\big)}{n!\;\Gamma\big(\tfrac{1}{4}\big)}=(-1)^n\binom{\small{-1/4}}{n} $$where \(\binom{n}{k}\) is the binomial coefficient though I haven't yet figured out the family that this example belongs to.

Entry 27

Here are some Ramanujan-inspired formulas for pi using the complete elliptic integral of the first kind \(K(k_d)\) and the golden ratio \(\phi=\tfrac{1+\sqrt{5}}{2}\)$$\frac{1}{\pi} = \frac{1}{2\,K(k_{5})} \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^6\phi^{6})^n}}$$ $$\frac{1}{\pi} = \frac{1}{2\,K(k_{15})} \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^{12}\phi^{8})^n}}$$ $$\frac{1}{\pi} = \frac{1}{2\,K(k_{25})} \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^6\phi^{24})^n}}$$ The denominators having different powers of \(\phi\) are the exact values of \(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}\) for \(\tau=\frac{1+\sqrt{-5}}{2},\,\frac{1+\sqrt{-15}}{2},\,\frac{1+\sqrt{-25}}{2},\) respectively. (Using \(d = -35\) would already need a sextic.) The last implies $$e^{\pi\sqrt{25}} \approx 2^6 \phi^{24}-24.00004\dots$$ And \(K(k_n)\) can be expressed as a product of gamma functions as given in the link above. For example $$K(k_5)=\frac{\phi^{3/4}\sqrt{\Gamma(\frac{1}{20})\,\Gamma(\frac{3}{20})\,\Gamma(\frac{7}{20})\,\Gamma(\frac{9}{20})}}{4\sqrt{10\pi}}=1.576390\dots$$ which can also be numerically evaluated by Wolfram Alpha.

Entry 26

Watson's triple integral \(I_1\) is$$\begin{aligned}I_1 &= \frac{1}{\pi^3}\int_0^\pi \int_0^\pi \int_0^\pi \frac{dx\, dy\, dz}{1-\cos x\cos y\cos z}\\
&=\left( \frac{2\sqrt{2}}{\pi}\int_0^{\pi/2}  \frac{dx}{\sqrt{1+\sin^2 x}}\right)^2\\
&= \frac{\Gamma^4(\frac{1}{4})}{4\pi^3}=4\,\eta(i)^4 = 1.393203\dots
\end{aligned}$$where \(\eta(\tau)\) is the Dedekind eta function and which was discussed in Entry 25. I found a nice Ramanujan/Chudnovsky-type formula for \(I_1\) using the golden ratio \(\phi=\tfrac{1+\sqrt{5}}{2}\) $$I_1 =\frac{25\phi^6}{\sqrt{\phi^{24}-4}}\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \left(\frac{-\phi^{16}}{4(\phi^{24}-4)}\right)^{3n}=1.393203\dots$$Notice that the \(24\)th power of the golden ratio is off by \(4\).

Entry 25

Let \(\eta(\tau)\) be the Dedekind eta function. In his Lost Notebook, Ramanujan came up with the nice evaluations$$\begin{aligned}\eta(i) &= \frac{\Gamma\big(\tfrac{1}{4}\big)}{2\,\pi^{3/4}}\\\eta(2i) &= \frac{\eta(i)}{2^{3/8}}\\ \eta(3i) &= \frac{\eta(i)}{3^{3/8}\,(2+\sqrt{3})^{1/12}} \\ \eta(4i) &= \frac{\eta(i)}{2^{13/16}\,(1+\sqrt{2})^{1/4}} \end{aligned}$$Extending this list, we find $$\begin{aligned}\eta(5i) &= \frac{\eta(i)}{\sqrt{5}}\left(\tfrac{1+\sqrt{5}}{2}\right)^{-1/2} \\\eta(6i) &=\frac{\eta(i)}{6^{3/8}} \Big(\tfrac{5-\sqrt{3}}{2}-\tfrac{3^{3/4}}{\sqrt{2}}\Big)^{1/6} \\ \eta(7i) &= \frac{\eta(i)}{\sqrt{7}}\left(-\tfrac{7}{2}+\sqrt{7}+\tfrac{1}{2}\sqrt{-7+4\sqrt{7}} \right)^{{1/4}}\\ \eta(8i) &= \frac{\eta(i)}{2^{41/32}} \frac{(-1+\sqrt[4]{2})^{1/2}}{(1+\sqrt{2})^{1/8}}\\ \eta(16i) &= \frac{\eta(i)}{2^{113/64}} \frac{(-1+\sqrt[4]{2})^{1/4}}{(1+\sqrt{2})^{1/16}} \left(-2^{5/8}+\sqrt{1+\sqrt{2}}\right)^{1/2}\end{aligned}$$It seems that for prime \(p=4m+1\), then \(x=\Big(\sqrt{p}\,\tfrac{\eta(p\,i)}{\eta(i)}\Big)^\color{blue}2\) is an algebraic number of degree \(\frac{p-1}{2}\), while if \(p=4m+3\) for \(p>3\), then \(y=\Big(\sqrt{p}\,\tfrac{\eta(p\,i)}{\eta(i)}\Big)^\color{blue}4\)  has degree \(\frac{p+1}{2}\). For the special case \(p=7\), note that $$y =\left(-\tfrac{7}{2}+\sqrt{7}+\tfrac{1}{2}\sqrt{-7+4\sqrt{7}} \right)\approx0.092192$$is one of the four quartic roots such that the polynomial on the \(\text{RHS}\) vanishes $$\frac{(y^2 + 5y + 1)^3(y^2 + 13y + 49)}{y} = 12^3+\frac{(y^4 + 14y^3 + 63y^2 + 70y - 7)^2}{y}$$ and the \(\text{LHS}\) assumes the value \(12^3=1728\).

Entry 24

There is this curious identity$$\sqrt{2\,\Big(1-\frac{1}{3^2}\Big) \Big(1-\frac{1}{7^2}\Big)\Big(1-\frac{1}{11^2}\Big)\Big(1-\frac{1}{19^2}\Big)} = \Big(1+\frac{1}{7}\Big)\Big(1+\frac{1}{11}\Big)\Big(1+\frac{1}{19}\Big)$$ Berndt, the primary editor of Ramanujan's Notebooks, asks: if this is an isolated result, or are there others? After a quick session with Mathematica, it turns out we can use the rest of the primes within that range $$\sqrt{2\,\Big(1-\frac{1}{2^6}\Big) \Big(1-\frac{1}{5^2}\Big)\Big(1-\frac{1}{13^2}\Big)\Big(1-\frac{1}{17^2}\Big)} = \Big(1+\frac{1}{5}\Big)\Big(1+\frac{1}{13}\Big)\Big(1+\frac{1}{17}\Big)$$ It seems far-fetched that he randomly scribbled fractions with prime denominators and just so happen to observe the equality. But with the description that "Every positive integer is one of Ramanujan's personal friends" (quote by Littlewood), then that might be just how he did it.

Entry 23

For powers \(k=4n+3\), Ramanujan gave $$\sum_{n=1} \frac{n^{3}}{e^{2n\pi}-1} = \frac{1^{3}}{e^{2\pi}-1}+\frac{2^{3}}{e^{4\pi}-1}+\frac{3^{3}}{e^{6\pi}-1}+\cdots=\frac{\Gamma\big(\tfrac{1}{4}\big)^8}{2^{10}\cdot5\pi^6}-\frac{1}{240}$$ $$\sum_{n=1} \frac{n^{7}}{e^{2n\pi}-1} = \frac{1^{7}}{e^{2\pi}-1}+\frac{2^{7}}{e^{4\pi}-1}+\frac{3^{7}}{e^{6\pi}-1}+\cdots=\frac{3\,\Gamma\big(\tfrac{1}{4}\big)^{16}}{2^{17}\cdot5\pi^{12}}-\frac{1}{480}$$ $$\sum_{n=1}^\infty  \frac{n^{11}}{e^{2n\pi}-1} = \frac{1^{11}}{e^{2\pi}-1}+\frac{2^{11}}{e^{4\pi}-1}+\frac{3^{11}}{e^{6\pi}-1}+\cdots =\frac{189\,\Gamma\big(\tfrac{1}{4}\big)^{24}}{2^{22}\cdot5\cdot13\,\pi^{18}}-\frac{691}{65520}$$ while for \(k=4n+1\) it evaluates to a rational number $$\sum_{n=1} \frac{n^{5}}{e^{2n\pi}-1} = \frac{1^{5}}{e^{2\pi}-1}+\frac{2^{5}}{e^{4\pi}-1}+\frac{3^{5}}{e^{6\pi}-1}+\cdots=\frac{1}{504}$$ $$\sum_{n=1} \frac{n^{9}}{e^{2n\pi}-1}=\frac{1^{9}}{e^{2\pi}-1}+\frac{2^{9}}{e^{4\pi}-1}+\frac{3^{9}}{e^{6\pi}-1}+\cdots=\frac{1}{264}$$ $$\sum_{n=1} \frac{n^{13}}{e^{2n\pi}-1}=\frac{1^{13}}{e^{2\pi}-1}+\frac{2^{13}}{e^{4\pi}-1}+\frac{3^{13}}{e^{6\pi}-1}+\cdots=\frac{1}{24}$$ and so on. The case \(k=1\) diverges.

Entry 22

The Ramanujan tau function \(\tau(n)\) is given by the coefficients of the q-expansion of the Dedekind eta function \(\eta(z)\)'s \(24\)th power.  Let \(q = e^{2\pi i\,z}\), then$$\begin{aligned}\eta(z)^{24} &= \sum_{n=1}^\infty\tau(n)q^n\\&=q - 24q^2 + 252q^3 - 1472q^4 + 4830q^5 - 6048q^6 - 16744q^7 + \dots\end{aligned}$$Let \(n\) be a prime \(p\). Ramanujan observed the remarkable congruence$$\tau(p)-1-p^{11}\equiv 0\ \bmod\ 691$$For example$$\begin{aligned}-24-1-2^{11}&= -691\times3\\252-1-3^{11}&= -691\times256\\4830-1-5^{11}&= -691\times70656\\-16744-1-7^{11}&= -691\times2861568\end{aligned}$$and so on. More generally, what he observed was$$\tau(n)\equiv\sigma_{11}(n)\ \bmod\ 691$$where \(\sigma_k(n)\) is the sum of the \(k\)th powers of the divisors of \(n\).

Entry 21

This infinitely nested radical is the case \(a=2\) of Entry 20 (which was the case \(a=5\)). The patterns are the same, namely \(\small{(+,+,+,-),\; (+,+,-,+),\; (+,-,+,+),\; (-,+,+,+)}\) $$x_1=\small{+\sqrt{2+\sqrt{2+\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\color{red}-\cdots}}}}}}}} = 2\cos\tfrac{2\pi}{17}=1.8649\dots$$ $$x_2=\small{+\sqrt{2+\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\color{red}-\sqrt{2+\cdots}}}}}}}} = 2\cos\tfrac{4\pi}{17}=1.4780\dots$$ $$x_3=\small{+\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\cdots}}}}}}}} = 2\cos\tfrac{8\pi}{17}=0.1845\dots $$ $$x_4=\small{\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}}}}} = 2\cos\tfrac{16\pi}{17}=-1.9659\dots $$ The \(x_i\) are roots of a quartic (with coefficients in \(\sqrt{17}\)). This implies that the four roots obey the system$$\begin{aligned}
x_1^2 &= x_2+2\\ x_2^2 &= x_3+2\\ x_3^2 &= x_4+2\\ x_4^2 &= x_1+2\\
\end{aligned}$$ For positive integer \(a\), apparently only \(a=2,5\) yield \(x_i\) that do not have cubic irrationalities.

Entry 20

One of Ramanujan’s talents was spotting elegant special cases of general phenomena. Consider for example$$x=\small+\sqrt{a+\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a-\cdots}}}}}}}\tag1$$where the repeating pattern of the signs is \((+,+,+,-)\). Generally, \(x\) will be the root of a \(12\)th deg factor of a \(16\)th deg equation. But he found that $$x_1=\small+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}=2.7472\dots$$ Note that \(x\) is just a root of a quartic. Its other roots are given by the patterns \((+,+,-,+),\; (+,-,+,+),\; (-,+,+,+)\), respectively $$x_2=\small+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\cdots}}}}}}} = \frac{2-\sqrt 5 +\sqrt{15+6\sqrt 5}}{2}=2.5473\dots$$ $$x_3=\small+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\cdots}}}}}}} = \frac{2+\sqrt 5 -\sqrt{15-6\sqrt 5}}{2}=1.4888\dots$$ $$x_4=\small\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\cdots}}}}}}} = \frac{2-\sqrt 5 -\sqrt{15+6\sqrt 5}}{2}=-2.7833\dots$$ This immediately implies that the four roots obey the system with \(a=5\), $$\begin{aligned}
x_1^2 &= x_2+a\\ x_2^2 &= x_3+a\\ x_3^2 &= x_4+a\\ x_4^2 &= x_1+a\\
\end{aligned}\tag2$$also studied by Ramanujan. In general, an infinitely nested radical with period length of 4 like \((1)\)$$x = \pm\sqrt{a\pm \sqrt{a\pm \sqrt{a\pm \sqrt{a\pm\dots}}}}$$and a system of 4 equations like \((2)\) can be expressed as$$x = (((x^2 - a)^2 - a)^2 - a)^2 - a\tag3$$Expanded out and factored, Ramanujan stated that \((3)\) was a product of 4 quartics, three of which had coefficients in the cubic,$$y^3+3y = 4(1+ay)$$However, this has a rational factor for the special cases when \(a=2,5\) so explains why the radical he found has no cubic irrationalities.

Entry 19

The value of the infinitely nested radical$$F(m)=\sqrt[m]{1+\sqrt[m]{1+\sqrt[m]{1+\sqrt[m]{1+\dots}}}}$$for integer \(m>1\) is well-known to be an algebraic number of degree \(m\). For \(m=2\), it is the golden ratio while \(m=3\) yields the plastic constant. Define$$G(n)=\sqrt[n]{1+2\sqrt[n]{1+3\sqrt[n]{1+4\sqrt[n]{1+\dots}}}}$$For degree \(n=2\), Ramanujan found that it was simply$$3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\dots}}}}$$More generally$$x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}$$ To see this, note that$$x+1=\sqrt { \left( x+1 \right) ^{2}}=\sqrt {1+{x}^{2}+2\,x}=\sqrt {1+x\sqrt {(x+2)^2}}\\=\sqrt {1+x\sqrt {1+{x}^{2}+4\,x+3}}=\sqrt {1+x\sqrt {1+ \left( x+1 \right) \sqrt {(x+3)^2}}} =\dots$$However, closed-forms for higher \(n\) such as \(G(3) \approx 1.702219132695458\) are not known.

Entry 18

Ramanujan had several expressions for the Riemann zeta function  $$\zeta(s) = \sum_{k=1}^\infty \frac{1}{k^s}$$ especially for \(s=3\), one of which was a rediscovery of Euler's continued fraction (useful for any general series) $$\beta\,\zeta(3) = \cfrac{1}{u_1-\cfrac{1^6}{u_2-\cfrac{2^6}{u_3-\cfrac{3^6}{u_4-\ddots}}}}$$where \(\beta = 1\) and the \( u_n\), starting with \(n = 1\), are, $$u_n = \color{brown}{(n-1)^3+n^3} = (2n-1)(n^2-n+1) = 1, 9, 35, 91, \dots$$Apery found an accelerated version, $$u_n = \color{brown}{n^3 + (n-1)^3 + 4(2n-1)^3} = (2n-1)(17n^2-17n+5) = 5, 117, 535, 1463, \dots$$(notice the sum of cubes) where now \(\beta = \frac{1}{6}\), and established that its rate of convergence was such that \(\zeta(3)\) could not be a ratio of two integers, thus famously proving its irrationality. Ramanujan also gave,$$ \zeta(3) = 1+\cfrac{1}{v_1+\cfrac{1^3}{1+\cfrac{1^3}{v_2+\cfrac{2^3}{1+\cfrac{2^3}{v_3 + \ddots}}}}}$$where the \(v_n\), again starting with \(n = 1\), are given by the linear function,$$ v_n = 4(2n-1) = 4, 12, 20, 28, \dots$$Using an approach similar to Apery's of finding a faster converging version, I found via Mathematica that,$$ \zeta(3) = \cfrac{6}{w_1 + \cfrac{1^3}{1 + \cfrac{1^3}{w_2 + \cfrac{2^3}{1 + \cfrac{2^3}{w_3 +\ddots}}}}}$$where the \(w_n\) are now defined by the cubic function, $$w_n = 4(2n-1)^3 = 4, 108, 500, 1372, \dots$$ Unfortunately, no comparable continued fraction is yet known that proves the irrationality of \(\zeta(5)\).

Entry 17

Here is another family (now based on \(d=-10\)) of pi approximations with a form consistent to the one (based on \(d=-58\)) in Entry 16. Define the fundamental units $$\begin{aligned}
U_{2} &= 1+\sqrt{2}\\
U_{5} &= \frac{1+\sqrt{5}}{2}\\
U_{10} &= 3+\sqrt{10}\\
U_{30} &= 11+2\sqrt{30}\end{aligned}$$ Then with increasing precision, accurate to \(5,10,14,18\) digits $$\begin{aligned}
\pi &\approx \frac{1}{\sqrt{10}} \ln \Big[ 2^6 (U_{5})^{12} \Big]\\
\pi &\approx \frac{1}{2\sqrt{10}} \ln \left[ 2^9 \left(U_2\, U_{5} \sqrt{U_{10}} \,\right)^6 \right]\\
\pi &\approx \frac{1}{3\sqrt{10}} \ln \left[ 2^6 (U_{5})^{12}\, (U_{30})^2 \left( \sqrt{\frac{3+\sqrt{6}}{4} } + \sqrt{\frac{-1+\sqrt{6}}{4}}\right)^{24}\right]\\
\pi &\approx \frac{1}{4\sqrt{10}} \ln \Big[ 2^9 \left(U_2\, U_{5} \sqrt{2\,U_{10}} \,\right)^3 \big(\sqrt{v+1} +\sqrt{v}\big)^{12}\Big]\end{aligned}$$ where \(v = 2^{-1/2}(U_2)^2\,(U_{5})^3\).

If I remember correctly, I think Ramanujan found the first two members. The expression inside the log function is again the exact value of \(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}\) where  \(\eta(\tau)\) is the Dedekind eta function, and  \(\tau = \frac{\sqrt{-10}}{2},  \tau = \frac{2\sqrt{-10}}{2},  \tau = \frac{3\sqrt{-10}}{2},  \tau = \frac{4\sqrt{-10}}{2}\), respectively.
 

Entry 16

Ramanujan gave an unusual approximation to \(\pi\)
$$\pi \approx \frac{4}{\sqrt{522}}\ln\left[\Big(\frac{5+\sqrt{29}}{\sqrt{2}}\Big)^3(5\sqrt{29}+11\sqrt{6})\left(\sqrt{\frac{9+3\sqrt{6}}{4}}+\sqrt{\frac{5+3\sqrt{6}}{4}} \right)^6\right]$$ which is good to \(31\) digits. This in fact belongs to a family. First, define the fundamental units $$\begin{aligned}
U_{2} &= 1+\sqrt{2}\\
U_{29} &= \frac{5+\sqrt{29}}{2}\\
U_{58} &= 99+13\sqrt{58}\\
U_{174} &= 1451+110\sqrt{174}\end{aligned}$$ Note that \(174 = 6\times29\). These are involved in fundamental solutions to Pell equations.  For example, for \(x^2-58y^2 = -1\), it is \((x, y) = (99, 13)\), (see the values for \(U_{58}\)). Then with increasing precision $$\begin{aligned}
\pi &\approx \frac{1}{\sqrt{58}} \ln \Big[ 2^6 (U_{29})^{12} \Big]\\
\pi &\approx \frac{1}{2\sqrt{58}} \ln \left[ 2^9 \left((U_2)^3\, U_{29} \sqrt{U_{58}} \,\right)^6 \right]\\
\pi &\approx \frac{1}{3\sqrt{58}} \ln \left[ 2^6 (U_{29})^{12}\, (U_{174})^2 \left( \sqrt{\frac{9+3\sqrt{6}}{4} } + \sqrt{\frac{5+3\sqrt{6}}{4}}\right)^{24}\right]\\
\pi &\approx \frac{1}{4\sqrt{58}} \ln \Big[ 2^9 \left((U_2)^3\, U_{29} \sqrt{2\,U_{58}} \,\right)^3 \big(\sqrt{v+1} +\sqrt{v}\big)^{12}\Big]\end{aligned}$$ where \(v = 2^{-1/2}(U_2)^6\,(U_{29})^3\).

Beautifully consistent, aren't they?  The last is by this author and is accurate to \(42\) digits. The expression inside the log function is the exact value of \(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}\) where  \(\eta(\tau)\) is the Dedekind eta function, and  \(\tau = \frac{\sqrt{-58}}{2},  \tau = \frac{2\sqrt{-58}}{2},  \tau = \frac{3\sqrt{-58}}{2},  \tau = \frac{4\sqrt{-58}}{2}\), respectively.

Anyone can find a nice expression for the next step?  

Entry 15

Ramanujan gave the following \(4\)th power identities
$$(2x^2+12xy-6y^2)^4+(2x^2-12xy-6y^2)^4+(4x^2-12y^2)^4+(4x^2+12y^2)^4+(3x^2+9y^2)^4=(5x^2+15y^2)^4$$ $$(6x^2-44xy-18y^2)^4+(8x^2+40xy-24y^2)^4+(14x^2-4xy-42y^2)^4+(9x^2+27y^2)^4+(4x^2+12y^2)^4=(15x^2+45y^2)^4$$ Expressed as the sextuple $$z_1^4+z_2^4+z_3^4+z_4^4+z_5^4=z_6^4$$ note that the two obey \(z_1+z_2=z_3\). It can be shown that, just like its \(3\)rd power counterpart discussed in Entry 13, there are infinitely many such formulas. Use the identity (by yours truly) $$(ax^2+2u_1xy-3ay^2)^k+(bx^2-2u_2xy-3by^2)^k+\big((a+b)x^2-2u_3xy-3(a+b)y^2\big)^k = \big(a^k+b^k+(a+b)^k\big)(x^2+3y^2)^k$$ for \(k = 2,4\) and where \(u_1 =a+2b,\,u_2=2a+b,\,u_3=a-b\). Thus all one needs is to find an initial sextuple of form \(z_1+z_2=z_3\) like $$6^4+8^4+(6+8)^4+9^4+4^4=15^4$$ $$2^4+44^4+(2+44)^4+39^4+52^4=65^4$$ and distributing the \(\text{RHS}\) of the identity will yield a quadratic parameterization.

Entry 14

Ramanujan's sum of cubes identity is defined by the generating functions, $$\begin{aligned}
\sum_{n=0}^\infty a_n x^n &= \frac{1+53x+9x^2}{R_1}\\
\sum_{n=0}^\infty b_n x^n &= \frac{2-26x-12x^2}{R_1}\\
\sum_{n=0}^\infty c_n x^n &= \frac{2+8x-10x^2}{R_1}\\
\end{aligned}\tag{1}$$ where \(R_1 = 1-82x-82x^2+x^3\). Then $$a_n^3+b_n^3 = c_n^3 + (-1)^n\tag{2}$$ It turns out the \(a_n, b_n, c_n\) can also be expressed as $$\begin{aligned}
a_n &=-9p^2+176pq-851q^2 = 1,135,11151,\dots\\
b_n &=4(3p^2-56pq+263q^2) = 2,138,11468,\dots\\
c_n &=2(5p^2-90pq+409q^2) = 2,172,14258,\dots\\
d_n &= -(p^2-85q^2) = 1, -1,1,-1\dots
\end{aligned}\tag{3}$$ with \(p,q\) chosen to satisfy the Pell equation \(p^2-85q^2 =\mp 1\). (Actually, \(p,q\) are half-integers since one can use \(p^2-85q^2 =\mp 4\).) But Entry 13 shows there are infinitely many quadratic parametrizations to \((2)\). Thus we can find similar generating functions such as $$\begin{aligned}
\sum_{n=0}^\infty a_n x^n &= \frac{-9(417-5602x+x^2)}{R_2}\\
\sum_{n=0}^\infty b_n x^n &= \frac{8(-566-11315x+x^2)}{R_2}\\
\sum_{n=0}^\infty c_n x^n &= \frac{-6(877+6898x+x^2)}{R_2}\\
\end{aligned}\tag{4}$$ where \(R_2 = -1+184899x-184899x^2+x^3\). Then $$a_n^3+b_n^3 = c_n^3 + 1\tag{5}$$ and its quadratic parameterization $$\begin{aligned}
a_n &=3(3p^2-104pq+909q^2) = 3753, 693875529,\dots\\
b_n &=-2(4p^2-135pq+1119q^2) = 4528, 837313192,\dots\\
c_n &=6(p^2-37pq+348q^2) = 5262, 972979926,\dots\\
d_n &=p^2-321q^2  = 1,1,1,\dots
\end{aligned}\tag{6}$$ with \(p,q\) chosen to satisfy \(p^2-321q^2 = 1\). So there are infinitely many sum of cubes identity analogous to Ramanujan's.

Entry 13

In Question 441 of the Journal of the Indian Mathematical Society (JIMS), Ramanujan asked,

"Show that \( (3x^2+5xy-5y^2)^3 +(4x^2-4xy+6y^2)^3 +(5x^2-5xy-3y^2)^3 =(6x^2-4xy+4y^2)^3\) and find other quadratic expressions satisfying similar relations."

There are in fact infinitely many such quadratic expressions. Use the identity (by yours truly) $$(ax^2-v_1xy+bwy^2)^3 + (bx^2+v_1xy+awy^2)^3 + (cx^2+v_2xy+dwy^2)^3 + (dx^2-v_2xy+cwy^2)^3 \\= (a^3+b^3+c^3+d^3)(x^2+wy^2)^3$$ where \(v_1=c^2-d^2,\; v_2=a^2-b^2,\,\) and \(w=(a+b)(c+d)\). Thus all we need is an initial solution to \(a^3+b^3+c^3+d^3=0\) and the identity guarantees an infinite more. 

Entry 12

Ramanujan gave a beautiful Diophantine identity. Let \(ad=bc\). Then
$$\small64\big((a+b+c)^6+(b+c+d)^6-(c+d+a)^6-(d+a+b)^6+(a-d)^6-(b-c)^6\big)\times\\\small\big((a+b+c)^{10}+(b+c+d)^{10}-(c+d+a)^{10}-(d+a+b)^{10}+(a-d)^{10}-(b-c)^{10}\big)=\\\small45\big((a+b+c)^8+(b+c+d)^8-(c+d+a)^8-(d+a+b)^8+(a-d)^8-(b-c)^8\big)^2$$ known as the 6-10-8 Identity. This is a remarkable relationship by most people’s standards. In the words of Berndt, primary editor of Ramanujan’s Notebooks, “it is one of the most fascinating identities we have ever seen”. One can easily appreciate its simplicity of form and the rather unexpected use of high exponents. However, we can generalize this.

I. Define \(F_n = x_1^n+x_2^n+x_3^n-(y_1^n+y_2^n+y_3^n),\,\) where \(\,\small x_1+x_2+x_3=y_1+y_2+y_3 = 0.\,\) If \(F_2 = F_4 = 0\), then $$64F_6F_{10} = 45F_8^2\quad \text{(Ramanujan)}$$ $$25F_3F_{7} = 21F_5^2\quad \text{(Hirschhorn)}$$ II. Define \(F_n = x_1^n+x_2^n+x_3^n+x_4^n-(y_1^n+y_2^n+y_3^n+y_4^n),\,\) where also \(\,\small \sum x_i =\sum y_i= 0.\,\) If \(F_1 = F_3 = F_5 = 0\), then $$7F_4F_9 = 12F_6F_7\quad \text{(yours truly)}$$ For example \(x_i = (21,\,9,\,-13,\,-17)\) and \(y_i = (23,\,1,\,-3,\,-21)\). And so on for similar relations for higher powers.

Entry 11

The previous entry featured the Chudnovsky algorithm which uses the negative discriminant \(d=-163\) and has class number \(h(d)=1\). Just like Ramanujan's well-known formula which uses \(d=-4\times58\) and has \(h(d)=2\), the algorithm is also connected to Pell equations. Let \(n=3\times163=489\). The fundamental solution to $$u^2-489v^2 = 1$$ is then given by the expansion of the fundamental unit $$U_{n} = \left(\tfrac{1}{18}(\color{brown}{640320}-6)\sqrt{3}+4826\sqrt{163}\right)^2=u+v\sqrt{489}$$ Recall that \(e^{\pi\sqrt{163}}\approx\color{brown}{640320}^3+743.99999999999925.\) Some experimentation yielded $$3\sqrt{3}\big(U_n^{1/2}-U_n^{-1/2}\big)+18 = \color{brown}{2^2\cdot3^3\cdot7^2\cdot11^2}$$ $$\sqrt{163}\big(U_n^{1/2}+U_n^{-1/2}\big) = \color{brown}{2^2\cdot19\cdot127\cdot163}$$ $$3\sqrt{3}\big(U_n^{1/2}-U_n^{-1/2}\big)+6 = \color{brown}{640320}$$ and we find these integers in the Chudnovsky algorithm $$12\sum_{k=0}^\infty (-1)^k \frac{(6k)!}{k!^3(3k)!}  \frac{\color{brown}{2\cdot3^2\cdot7\cdot11\cdot19\cdot127\cdot163}\,k+13591409}{(\color{brown}{640320}^3)^{k+1/2}} = \frac{1}{\pi}$$ The same thing happens with the other large \(d\) with \(h(d)=1\), for example \(3\times67=201\) $$U_{201} = \left(\tfrac{1}{18}(\color{brown}{5280}-6)\sqrt{3}+62\sqrt{67}\right)^2$$ and \(e^{\pi\sqrt{67}}\approx\color{brown}{5280}^3+743.9999986.\) Similar expressions using \(U_{201}\) will yield the integers for its corresponding pi formula.

Entry 10

The Chudnovsky algorithm is $$\frac{1}{\pi} = 12 \sum^\infty_{k=0} (-1)^k\frac{ (6k)!}{k!^3\,(3k)!}\frac{ (163 \times 3344418 k + 13591409)}{ (640320^3)^{k + 1/2}}\tag1$$ Recall the famous $$e^{\pi\sqrt{163}}=640320^3+743.99999999999925\dots$$ The formula \((1)\) was inspired by Ramanujan's work and is used to calculate world records for the digits of pi. Now why didn't he discover this? Actually, he almost did. In his list of 17 formulas, there were two that belong to this family, one of which is $$\frac{1}{\pi} = \frac{18\sqrt{3}}{85\sqrt{85}} \sum_{n=0}^{\infty} \frac{\big(\tfrac{1}{2}\big)_n\big(\tfrac{1}{6}\big)_n\big(\tfrac{5}{6}\big)_n}{n!^3}\, (133n+8)\,\Big(\frac{4}{85}\Big)^{3n}\tag2$$ where \((a)_n\) is a Pochhammer symbol. However, this can be translated into the form of \((1)\) namely $$\frac{1}{\pi} = 162 \sum^\infty_{k=0} \frac{ (6k)!}{k!^3\,(3k)!}\frac{ (133 k + 8)}{ (255^3)^{k + 1/2}}\tag3$$ Similarly,$$e^{\pi\sqrt{28}}=255^3-744.01\dots$$ where the "excess" \(744\) indicates that the j-function is involved. In fact, in Ramanujan's Lost Notebook, he had calculations involving Eisenstein series using the primes \(d=11,19,43,67,163\) which is precisely the family which \((1)\) belongs to. Thus, if only Ramanujan lived longer, he would almost have surely found the Chudnovsky algorithm.

Entry 9

Here is another "bizarre" continued fraction from Ramanujan involving \(e\) and \(\pi\). For \(x>0\) $$\sqrt{\frac{\pi\,e^x}{2x}}=1+\frac{x}{1\cdot3}+\frac{x^2}{1\cdot3\cdot5}+\frac{x^3}{1\cdot3\cdot5\cdot7}+\dots+\cfrac1{x+\cfrac{1}{1+\cfrac{2}{x+\cfrac{3}{1+\ddots}}}}$$As Kevin Brown of Mathpages commented in this old sci.math post, "Is there any other mathematician whose work is instantly recognizable?" Note that the error function has a reminiscent form (also rediscovered by Ramanujan) $$\int_0^x e^{-t^2}dt=\tfrac{1}{2}\sqrt{\pi}\,\text{erf}(x)=\tfrac{1}{2}\sqrt{\pi}-\cfrac1{x+\cfrac{1}{2x+\cfrac{2}{x+\cfrac{3}{2x+\ddots}}}}$$

Entry 8

The previous entry ended with \(\cos\big(\tfrac{2\pi}{29}\big)\). This is not by Ramanujan, but I got reminded about something elegant regarding \(p=29\) which he would have appreciated. Does anyone know why the octic found by Igor Schein in 1999 $$x^8-x^7+29x^2+29 = 0$$is solvable in radicals, specifically by the \(29\)th root of unity? Any other octic or higher with a similarly simple form?

Incidentally, the discriminant of \(x^8-x^7+ax^2+a\) is $$F(a)=186624 a^3 - 3561092 a^2 + 29511140 a - 7^7$$ and the only integer solution to the elliptic curve \(F(a) = y^2\) is \(a = 29\).

Entry 7

Ramanujan found the nice trigonometric relation, $$\sqrt[3]{2\cos\big(\tfrac{2\pi}{7}\big)}+\sqrt[3]{2\cos\big(\tfrac{4\pi}{7}\big)} +\sqrt[3]{2\cos\big(\tfrac{6\pi}{7}\big)}  = -\sqrt[3]{-5+3\cdot7^{1/3}}= -0.904\dots$$ Equivalently, let \(x_i\) be the three roots of the cubic \(x^3+x^2-2x-1=0\). Then $$\sum_{k=1}^3 x_k^{1/3} = -\sqrt[3]{-5+3\cdot7^{1/3}} = -0.904\dots$$ The natural question to ask is, can this be generalized to quintics and \(\cos\big(\tfrac{2\pi}{11}\big)\)? It turns out it can. Noam Elkies found that if we let \(y_i\) be the five roots of the quintic \(y^5+6y^4-y^3-32y^2+16y-1=0\), then $$\sum_{k=1}^5 y_k^{1/5} = -\sqrt[5]{-274+5(-21\cdot11^{1/5}+13\cdot11^{2/5}+13\cdot11^{3/5})}=-0.093\dots$$ The \(y_i\) can also be expressed as cosines $$y_k=-(z_k^2-1)^2-(z_k-1),\quad \text{and}\; z_k=2\cos\big(\tfrac{2\pi\,k}{11}\big)$$ Similarly, one can find a \(7\)th deg relation using \(\cos\big(\tfrac{2\pi}{29}\big)\) and so on.

Entry 6

Ramanujan was a master at manipulating radicals. Some of his unusual ones are
$$\sqrt[3]{\sqrt[5]{\frac{32}{5}} - \sqrt[5]{\frac{27}{5}}} = \sqrt[5]{\frac{1}{25}} + \sqrt[5]{\frac{3}{25}} + \sqrt[5]{\frac{-9}{25}}\tag1$$ $$\sqrt{\sqrt[5]{\frac{1}{5}} + \sqrt[5]{\frac{4}{5}}} = - \sqrt[5]{\frac{1}{125}}+\sqrt[5]{\frac{2}{125}} + \sqrt[5]{\frac{8}{125}} + \sqrt[5]{\frac{16}{125}} \tag2$$ $$\sqrt[4]{\frac{3 + 2\sqrt[4]{5}}{3 - 2\sqrt[4]{5}}} = \frac{\sqrt[4]{5} + 1}{\sqrt[4]{5} - 1}\tag3$$ $$\sqrt[8]{1+\sqrt{1-\left(\frac{-1+\sqrt{5}}{2}\right)^{24}}} = \frac{-1+\sqrt{5}}{2}\,\frac{1+\sqrt[4]{5}}{\sqrt{2}}\tag4$$ The pattern of \((2)\) is certainly suggestive, though I haven't seen any generalization of this. And \((4)\) is the \(8\)th root of an expression involving a \(24\)th power! Also, note that he only uses ratios of \(\sqrt{p^k}\) with the small primes \(p=2,3,5\). Either such simple relations are possible only for these, or Ramanujan found the low-hanging fruits and there are similar ones with prime \(p>5\).

Entry 5

Given the partition function \(p(n)\), Ramanujan found,
$$\begin{align}
p(5k+4) & \equiv 0 \pmod 5 \\
p(7k+5) & \equiv 0 \pmod 7 \\
p(11k+6) & \equiv 0 \pmod {11}
\end{align}$$ It was later proved in 2000 by Ken Ono, building on the work of A.O. Atkin, that such congruences exist modulo every integer \(m\) coprime to \(6\). Some examples from the link above, $$p(13 \cdot 11^3k + 237)\equiv 0 \pmod {13}$$ $$p(19 \cdot 101^4k + 815655)\equiv 0 \pmod {19}$$ $$p(31\cdot 107^4k + 30064597)\equiv 0\pmod{31}$$ Given a Ramanujan-type congruence \(p(Ak+B) \equiv 0 \pmod m\) with \(m>11\). Must the factorization of \(A\) involve an integer with a power \(> 1\)? Also, I think the one with \(m=47\) should involve smaller integers than the last three examples.

Entry 4

These q-continued fractions were "missed(?)" by Ramanujan. Given the golden ratio \(\phi = \frac{1+\sqrt{5}}{2}\), then $$\frac{\sqrt{5}}{1+\phi^{-1}\big({-v}+\sqrt{1+v^2}\big)^{1/5}}-\phi = \cfrac{e^{-2\pi\sqrt{5}/5}}{1 + \cfrac{e^{-2\pi\sqrt{5}}}{1 + \cfrac{e^{-4\pi\sqrt{5}}}{1 + \cfrac{e^{-6\pi\sqrt{5}}}{1 + \ddots}}}}\quad\text{where}\; v = \phi^5\tag1$$ $$\frac{\sqrt{5}}{1+\phi^{-1}\big({-v}+\sqrt{1+v^2}\big)^{1/5}}-\phi = \cfrac{e^{-2\pi\sqrt{10}/5}}{1 + \cfrac{e^{-2\pi\sqrt{10}}}{1 + \cfrac{e^{-4\pi\sqrt{10}}}{1 + \cfrac{e^{-6\pi\sqrt{10}}}{1 + \ddots}}}}\quad\text{where}\; v = \small{18-5\sqrt{5}}\tag2$$ $$\frac{\sqrt{5}}{1+\phi^{-1}\big({-v}+\sqrt{1+v^2}\big)^{1/5}}-\phi = \cfrac{e^{-2\pi\sqrt{15}/5}}{1 + \cfrac{e^{-2\pi\sqrt{15}}}{1 + \cfrac{e^{-4\pi\sqrt{15}}}{1 + \cfrac{e^{-6\pi\sqrt{15}}}{1 + \ddots}}}}\quad\text{where}\; v = \tfrac{147-55\sqrt{5}}{4}\tag3$$ where, as \(\sqrt{5n}\) increases, then \(v\) is an algebraic number of generally increasing high degree. I found this family using Mathematica and the second cfrac in Entry 3. This implies that the Rogers-Ramanujan continued fraction \(R(\tau)\) $$R(i \sqrt{n})=\frac{\sqrt{5}}{1+\phi^{-1}R(i/\sqrt{n})}-\phi$$ for some positive real \(n\), though I have no proof of this assertion.

Entry 3

Ramanujan gave the beautiful continued fractions (with the second simplified by this author) $$5^{1/4}\sqrt{\phi}\,-\phi=\cfrac{e^{-2\pi/5}}{1 + \cfrac{e^{-2\pi}}{1 + \cfrac{e^{-4\pi}}{1 + \cfrac{e^{-6\pi}}{1 + \ddots}}}}\tag1$$ $$\frac{\sqrt{5}}{1+\phi^{-1}\big({-\phi^5}+\sqrt{\phi^{10}+1}\big)^{1/5}}-\phi=\cfrac{e^{-2\pi/\sqrt{5}}}{1 + \cfrac{e^{-2\pi\sqrt{5}}}{1 + \cfrac{e^{-4\pi\sqrt{5}}}{1 + \cfrac{e^{-6\pi\sqrt{5}}}{1 + \ddots}}}}\tag2$$ $$\big({-\phi^5}+\sqrt{\phi^{10}+1}\big)^{1/5}=\cfrac{e^{-2\pi/(5\sqrt{5})}}{1 + \cfrac{e^{-2\pi/\sqrt{5}}}{1 + \cfrac{e^{-4\pi/\sqrt{5}}}{1 + \cfrac{e^{-6\pi/\sqrt{5}}}{1 + \ddots}}}}\tag3$$with the golden ratio \(\phi = \frac{1+\sqrt{5}}{2}\). In 1913, the British mathematician G.H. Hardy, after reading the letter Ramanujan sent to him (which included examples of these extraordinary continued fractions), remarked, “…the [theorems] defeated me completely; I had never seen anything in the least like them before.” He would have been even more amazed had he known that these were connected to geometry.

Entry 2

Here's another pi formula from Ramanujan,$$\frac{1}{\pi} =\frac{2 \sqrt 2}{\color{blue}{9}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{5\cdot\color{blue}{2\cdot1}\,k+1}{\color{blue}{(12^4)}^k}$$ Note that $$e^{\pi\sqrt{10}} = 12^4-104.21\dots$$The approximation is not as impressive as in Entry 1, but shows the same pattern and "excess". It also has the same connection to Pell equations. Given the golden ratio (and fundamental unit) \(\phi = \frac{1+\sqrt{5}}{2}\), then $$\phi^3=2+\sqrt{5},\quad \text{thus}\;\;\color{blue}{2}^2-5\cdot\color{blue}{1}^2=-1$$ $$\phi^6=9+4\sqrt{5},\quad \text{thus}\;\;\color{blue}{9}^2-5\cdot4^2=1$$ $$2^6\left(\phi^6+\phi^{-6}\right)^2 =\color{blue}{12^4}$$ and, just like the previous entry, we find these integers all over the formula. However, there are only two negative fundamental discriminants of form \(d = -4(8n+2)\) with class number \(h(d)=2\), namely \(d = -4\times10\) and \(d=-4\times58\) so this nice direct connection to Pell equations, among Ramanujan's many pi formulas, is just limited to these two.

Entry 1

One of the most famous results of Ramanujan is his formula for \(\pi\) as $$\frac{1}{\pi} =\frac{2 \sqrt 2}{\color{blue}{9801}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot\color{blue}{70\cdot13}\,k+1103}{\color{blue}{(396^4)}^k}$$ Note that $$e^{\pi\sqrt{58}} = 396^4-104.00000017\dots$$ The terms have been factored to show some interesting connections to Pell equations. If we define the fundamental unit \(U_{29} = \frac{5+\sqrt{29}}{2}\), then
$$\big(U_{29}\big)^3=70+13\sqrt{29},\quad \text{thus}\;\;\color{blue}{70}^2-29\cdot\color{blue}{13}^2=-1$$ $$\big(U_{29}\big)^6=9801+1820\sqrt{29},\quad \text{thus}\;\;\color{blue}{9801}^2-29\cdot1820^2=1$$ $$2^6\left(\big(U_{29}\big)^6+\big(U_{29}\big)^{-6}\right)^2 =\color{blue}{396^4}$$ A similar situation happens with a pi formula using, of all things, the golden ratio. The fact that these are so has to do with the Dedekind eta function, but that's another story.