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“.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: