## Archive for the ‘geometry’ Category

### Hypergeometric formulas for Ramanujan’s continued fractions 2

(continued from yesterday’s post)

III. Icosahedral group

\begin{aligned} G(q) &= \sum_{n=0}^\infty \frac{q^{n^2}}{(q;q)_n} = \prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\\H(q) &= \sum_{n=0}^\infty \frac{q^{n^2+n}}{(q;q)_n} = \prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})}\end{aligned}

I observed that,

\begin{aligned}&q^{-1/60}G(q) = j^{1/60}\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j}\big) = (j-1728)^{1/60}\,_2F_1\big(\tfrac{29}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{1728-j}\big)\\[2.5mm]&q^{11/60}H(q) = j^{-11/60}\,_2F_1\big(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j}\big) = (j-1728)^{-11/60}\,_2F_1\big(\tfrac{41}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{1728-j}\big)\end{aligned}

where, as in the previous post, $j=j(\tau)$ is the j-function, $q = e^{2\pi i \tau} = \exp(2\pi i \tau)$, $\tau = \sqrt{-N}$, and $N>1$.  Since it is known that,

\begin{aligned}&r(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^2}{1 + \cfrac{q^3}{1 + \ddots}}}} = \frac{q^{11/60}H(q)}{q^{-1/60}G(q)} = \frac{q^{11/60}\prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})}}{q^{-1/60}\prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(q^{5n-4})}}\end{aligned}

this implies that,

\begin{aligned}r(q) &=\frac{j^{-11/60}\,_2F_1\big(\frac{31}{60},\frac{11}{60},\frac{6}{5},\frac{1728}{j}\big) }{j^{1/60}\,_2F_1\big(\frac{19}{60},\frac{-1}{60},\frac{4}{5},\frac{1728}{j}\big)}\\[3mm]&=\frac{(j-1728)^{-11/60}\,_2F_1\big(\frac{41}{60},\frac{11}{60},\frac{6}{5},\frac{1728}{1728-j}\big) }{(j-1728)^{1/60}\,_2F_1\big(\frac{29}{60},\frac{-1}{60},\frac{4}{5},\frac{1728}{1728-j}\big)}\end{aligned}

Example. Let $\tau = \sqrt{-4}$, hence $j = j(\sqrt{-4}) = 66^3$. Then,

$1/r(q) -r(q) = \left(\frac{1+\sqrt{5}}{2}\right)^4+\left(\frac{1+\sqrt{5}}{2}\right)5^{3/4} = 12.2643\dots$

Furthermore, since Ramanujan established that,

$G(q^{11})H(q)-q^2G(q)H(q^{11}) = 1$

if we define the two functions,

\begin{aligned}U(\tau) &= \big(j(\tau)\big)^{1/60}\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j(\tau)}\big)\\V(\tau) &= \big(j(\tau)\big)^{-11/60}\,_2F_1\big(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j(\tau)}\big) \end{aligned}

then the counterpart hypergeometric identity is also beautifully simple and given by,

$U(11\tau)V(\tau)-U(\tau)V(11\tau)=1$

In the next post, we will use one of the hypergeometric formulas to solve the general quintic.

### Hypergeometric formulas for Ramanujan’s continued fractions 1

There are five Platonic solids, two are duals to another two, while the tetrahedron is self-dual. As such, this gives rise to 3 polyhedral groups: the tetrahedral group of order 12, the octahedral group of order 24, and the icosahedral group of order 60.

Amazingly, Ramanujan found 3 continued fractions that can be associated with each group. See this article for more details.  It turns out there are also corresponding hypergeometric formulas, and the numbers 12, 24, and 60 naturally appears.

First though, define the j-function as,

$j = j(\tau) = \frac{1}{q} + 744 + 196884q + 21493760q^2 + \dots$

where,

$q = e^{2\pi i \tau} = \exp(2\pi i \tau)$

This can be conveniently calculated in Mathematica as,

$j(\tau) = 1728\text{KleinInvariantJ}(\tau)$

NOTE:  In the formulas below, it will be assumed that,

$\tau = \sqrt{-N},\;\; N > 1$

I. Tetrahedral group

Given,

\begin{aligned}&c=c(q)=\cfrac{q^{1/3}}{1 + \cfrac{q+q^2}{1 + \cfrac{q^2+q^4}{1 + \cfrac{q^3+q^6}{1 + \ddots}}}} = q^{1/3}\prod_{n=1}^\infty \frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})^2}\end{aligned}

and,

\begin{aligned}&d = j^{1/3}\,\frac{\,_2F_1\big(\frac{1}{4},\frac{-1}{12},\frac{2}{3},\frac{1728}{j}\big)}{\,_2F_1\big(\frac{1}{4},\frac{7}{12},\frac{4}{3},\frac{1728}{j}\big)}\end{aligned}

then we have the simple relationship,

$d = 4c^2+c^{-1}$

Example.  Let $\tau = \sqrt{-2}$, hence $j=j(\sqrt{-2}) = 20^3$, then,

$d = 3\sqrt{2(11+4\sqrt{6})} = 19.3484\dots$

and c(q) can then be easily solved for as a cubic equation.

II. Octahedral group

Let,

\begin{aligned}&u = u(q) = \cfrac{\sqrt{2}\,q^{1/8}}{1 + \cfrac{q}{1+q + \cfrac{q^2}{1+q^2 + \cfrac{q^3}{1+q^3 + \ddots}}}} = \sqrt{2}\,q^{1/8}\prod_{n=1}^\infty\frac{1-q^{2n-1}}{(1-q^{4n-2})^2} \end{aligned}

then,

\begin{aligned}&u = \frac{\sqrt{2}}{j^{1/8}}\left(\frac{\,_2F_1\big(\frac{5}{24},\frac{13}{24},\frac{5}{4},\frac{1728}{j}\big)}{\,_2F_1\big(\frac{7}{24},\frac{-1}{24},\frac{3}{4},\frac{1728}{j}\big)}\right)^{1/2}\end{aligned}

Example. Still using $\tau = \sqrt{-2}$, then,

$u = \sqrt{1+\sqrt{2}-\sqrt{2(1+\sqrt{2})}} = 0.4656\dots$

III. Icosahedral group

(To be discussed in the next post.)

### The Tremendous Tribonacci constant

The tribonacci constant is the real root of the cubic equation,

$T^3-T^2-T-1 = 0$

and is the limiting ratio of the tribonacci numbers = {0, 1, 1, 2, 4, 7, 13, 24, …} where each term is the sum of the previous three, analogous to the Fibonacci numbers.  Let $d = 11$, then,

$T = \frac{1}{3} +\frac{1}{3}\big(19+3\sqrt{3d}\big)^{1/3} + \frac{1}{3}\big(19-3\sqrt{3d}\big)^{1/3} = 1.839286\dots$

We’ll see that the tribonacci constant is connected to the complete elliptic integral of the first kind $K(k_{11})$.  But first, given the golden ratio’s infinite radical representation,

$\phi = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$

then T also has the beautiful infinite radical,

$\frac{1}{T-1} = \sqrt[3]{\frac{1}{2}+\sqrt[3]{\frac{1}{2}+\sqrt[3]{\frac{1}{2}+\sqrt[3]{\frac{1}{2}+\dots}}}} = 1.191487\dots$

as well as a continued fraction,

$\big(\frac{T}{T+1}\big) \big(e^{\frac{\pi\sqrt{11}}{24}}\big) = 1 + \cfrac{q}{1-q + \cfrac{q^3-q^2}{1+\cfrac{q^5-q^3}{1+\cfrac{q^7-q^4}{1+\ddots}}}} = 0.9999701\dots$

where q is the negative real number,

$q = \frac{-1}{e^{\pi \sqrt{11}}}$

Recall that at elliptic singular values, the complete elliptic integral of the first kind K(k) satisfies the equation,

$\frac{K'(k_d)}{K(k_d)} = \sqrt{d}$

or, in the syntax of Mathematica,

$\frac{EllipticK[1-ModularLambda[\sqrt{-d}]]}{EllipticK[ModularLamda[\sqrt{-d}]]} = \sqrt{d}$

Interestingly, we can express both $k_{11} = 0.000477\dots$ and $K(k_{11}) = 1.57098\dots$ in terms of the tribonacci constant as,

$k_{11} = \frac{1}{4}\left(2-\sqrt{\frac{2v+7}{2v-7}}\,\right) = 0.000477\dots$

where,

$v = T+4$

and,

$K(k_{11}) = \left(\frac{T+1}{T}\right)^2\, \frac{1}{11^{1/4}\, (4\pi)^{2}} \, \Gamma(\tfrac{1}{11}) \Gamma(\tfrac{3}{11}) \Gamma(\tfrac{4}{11}) \Gamma(\tfrac{5}{11}) \Gamma(\tfrac{9}{11})$

where $\Gamma(n)$ is the gamma function, as well as the infinite series,

$K(k_{11}) = \left(\frac{T+1}{T}\right)^2 \frac{\pi}{32^{1/4}} \sqrt{\frac{1}{4} \sum_{n=0}^\infty \frac{(6n)!}{(3n)!n!^3} \,\frac{1}{(-32)^{3n}} }$

With a slight tweak of the formula, we instead get,

$\frac{1}{4\pi} = \frac{1}{32^{3/2}} \sum_{n=0}^\infty \frac{(6n)!}{(3n)!n!^3}\, \frac{154n+15}{(-32)^{3n}}$

Finally, saving the best for last, given the snub cube, an Archimedean solid,

then the Cartesian coordinates for its vertices are all the even and odd permutations of,

{± 1,  ± 1/T,  ±}

with an even and odd number of plus signs, respectively, similar to how, for the vertices of the dodecahedron — a Platonic solid —  one can use the golden ratio.

For more about the tribonacci constant, and the equally fascinating plastic constant,  kindly refer to “A Tale of Four Constants “.

### Ramanujan’s pi approximations and Pell equations

Ramanujan gave many fascinating formulas and approximations to pi. Using one of his examples, we can give its family. First, define the fundamental units,

$U_{2} = 1+\sqrt{2}$

$U_{29} = \frac{5+\sqrt{29}}{2}$

$U_{58} = 99+13\sqrt{58}$

$U_{174} = 1451+110\sqrt{174}$

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 above), while for $x^2-174y^2 = 1$ it is {x, y} = {1451, 110}. Using these solutions to Pell equations, then,

$\pi \approx \frac{1}{\sqrt{58}} \ln \Big[ 2^6 (U_{29})^{12} \Big]$

$\pi \approx \frac{1}{2\sqrt{58}} \ln \left[ 2^9 \left((U_2)^3 U_{29} \sqrt{U_{58}} \,\right)^6 \right]$

$\pi \approx \frac{1}{3\sqrt{58}} \ln \left[ 2^6 (U_{29})^{12} (U_{174})^2 \left( \sqrt{\frac{9+3\sqrt{6}}{4} } + \sqrt{\frac{5+3\sqrt{6}}{4}}\right)^{24}\right]$

$\pi \approx \frac{1}{4\sqrt{58}} \ln \Big[ 2^9 \left((U_2)^3 U_{29} \sqrt{2U_{58}} \,\right)^3 \left(\sqrt{v+1} +\sqrt{v}\right)^{12}\Big]$

where,

$v = 2^{-1/2}(U_2)^6(U_{29})^3$

Nice, isn’t it?  The second to the last approximation is by Ramanujan which is accurate to 31 digits, while the last is by this author and is accurate to 42 digits.  (Can anyone find a nice expression for the next step? )  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.

The fundamental discriminant d = -4∙58 has class number h(d) = 2.  Another one with the same class number is d = -4∙37.  Hence, given,

$U_{37} = 6+\sqrt{37}$

$U_{111} = 295+28\sqrt{111}$

then,

$\pi \approx \frac{1}{\sqrt{37}} \ln \Big[ 2^6 (U_{37})^{6} \Big]$

$\pi \approx \frac{1}{3\sqrt{37}} \ln \left[ 2^6 (U_{37})^{6} (U_{111})^2 \left( \sqrt{\frac{37+20\sqrt{3}}{4} } + \sqrt{\frac{33+20\sqrt{3}}{4}}\right)^{12}\right]$

where the expression inside the log function is now the absolute value of the eta quotient at $\tau = \frac{1+\sqrt{-37}}{2}$ and $\tau = \frac{1+3\sqrt{-37}}{2}$.

### The Brioschi quintic and the Rogers-Ramanujan continued fraction

Given Ramanujan’s constant,

$e^{\pi\sqrt{163}} \approx Q+743.9999999999992\dots$

where $Q = 640320^3$, why do we know, in advance, that the quintic,

$w^5-10(\frac{1}{1728+Q})w^3+45(\frac{1}{1728+Q})^2w-(\frac{1}{1728+Q})^2 = 0$

is solvable in radicals?  The answer is this: The general quintic can be transformed in radicals to the one-parameter form,

$w^5-10cw^3+45c^2w-c^2 = 0$

called the Brioschi quintic.  Whether reducible or not, if it is solvable in radicals and c is rational, then it can be shown c must have the form,

$c = 1/(1728-t)$

where,

$t = \frac{(u^2+10u+5)^3}{u}$

for some radical u. (For example, u = 1 will yield an irreducible though solvable quintic.)  But it seems Nature likes to recycle polynomials as this is also one of the many formulas for the j-function $j(\tau)$, namely,

$j(\tau) = \frac{(v^2+10v+5)^3}{v}$

where,

$v = \Big(\frac{\sqrt{5}\,\eta(5\tau)}{\eta(\tau)}\Big)^6$

and $\eta(\tau)$ is the Dedekind eta function.  However, if we let,

$v = \frac{-125r^5}{r^{10}+11r^5-1}$

then we get the more well-known j-function formula,

$j(\tau) = \frac{-(r^{20}-288r^{15}+494r^{10}+228r^5+1)^3}{r^5(r^{10}+11r^5-1)^5}$

where the numerator and denominator are polynomial invariants of the icosahedron,

a Platonic solid wherein one can find pentagons (which, of course, has 5 sides).  Perhaps not surprisingly, r is given by the eta quotient,

$r^{-1}-r = \frac{\eta(\tau/5)}{\eta(5\tau)}+1$

But one can also express r using the beautiful Rogers-Ramanujan continued fraction. Let,

$q = e^{2 \pi i \tau}$

then,

$r = r(\tau) = \cfrac{q^{1/5}}{1+ \cfrac{q}{1 + \cfrac{q^2}{1+ \cfrac{q^3}{1 + \ddots}}}}$

One of the simplest cases is,

$r(\sqrt{-1}) = 5^{1/4}\sqrt{\frac{1+\sqrt{5}}{2}}-\frac{1+\sqrt{5}}{2}= \cfrac{e^{-2\pi/5}}{1+ \cfrac{e^{-2\pi}}{1 + \cfrac{e^{-4\pi}}{1+ \cfrac{e^{-6\pi}}{1 + \ddots}}}} \approx 0.284079$

which was communicated by Ramanujan to Hardy in his famous letter.

Interesting, isn’t it?

For a related topic, kindly read “Ramanujan’s Continued Fractions and the Platonic Solids“.