## Posts Tagged ‘icosahedron’

### Algebraic values of the Hypergeometric function

\begin{aligned} &\,_2F_1(a,b,c,z) = \sum_{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n} \frac{z^n}{n!}\end{aligned}

where $(a)_n$ is a Pochhammer symbol is, for rational {a, b, c, z}, generally not a rational value.  In the link given above, equations (18) and (19) are,

$\,_2F_1\big(\frac{1}{3},\frac{2}{3},\frac{5}{6},\frac{27}{32}\big) =\frac{8}{5}$

$\,_2F_1\big(\frac{1}{4},\frac{1}{2},\frac{3}{4},\frac{80}{81}\big) =\frac{9}{5}$

There are in fact an infinite number of such equalities.  One given by M. Glasser is, let $0 < x < \frac{1}{\sqrt{3}}$, then,

$\,_2F_1\big(\frac{1}{3},\frac{2}{3},\frac{3}{2},\frac{27x^2(1-x^2)^2}{4}\big) =\frac{1}{1-x^2}$

Another, based on eq. (42) of Vidunas’ “Transformations of algebraic Gauss hypergeometric functions” is, let $0 < y < 1$, then,

$\,_2F_1\big(\frac{1}{4},\frac{-1}{12},\frac{2}{3},\frac{(9-y^4)^3(-1+y^4)}{64y^{12}}\big) =\frac{1}{y}$

A third which yields not a rational but an algebraic number is remarkable for its connection to the Rogers-Ramanujan continued fraction. Let $0 < z < v_1$ where,

$v_1 = \text{Root}[z^4+228z^3+494z^2-228z+1=0] = 0.004428\dots$

then,

$\,_2F_1\big(\frac{19}{60},\frac{-1}{60},\frac{4}{5},\frac{1728z(z^2-11z-1)^5}{(z^4+228z^3+494z^2-228z+1)^3}\big) =\frac{1}{(z^4+228z^3+494z^2-228z+1)^{1/20}}$

If the polynomials are familiar, it is because they are invariants of the icosahedron.  They also appear in the j-function formula,

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

where,

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

and

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

### An Icosahedral and Brioschi quintic identity

Here’s an identity I found.  For arbitrary r, define,

$a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$

and,

$w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$

then,

$w^5-10aw^3+45a^2w-a^2 = 0$

Those two complicated expressions neatly wrap up into that last equation, doesn’t it?  This is the Brioschi quintic form which the general quintic can be reduced into.  Two of the polynomials are easily recognizable as icosahedral invariants, while,

$P(r) = r^6+2r^5-5r^4-5r^2-2r+1$

is a polynomial invariant for the octahedron. This gave rise to the question here.