## Posts Tagged ‘j-function’

### 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 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“.