Given Ramanujan’s constant,
where , why do we know, in advance, that the quintic,
is solvable in radicals? The answer is this: The general quintic can be transformed in radicals to the one-parameter form,
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,
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 , namely,
and is the Dedekind eta function. However, if we let,
then we get the more well-known j-function formula,
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,
But one can also express r using the beautiful Rogers-Ramanujan continued fraction. Let,
One of the simplest cases is,
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“.