Vtgyjfy

Given the Brioschi quintic

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

I'm interested in seeing different ways of solving it in terms of elliptic functions or theta functions.

To solve the general quintic using elliptic functions, one way is to reduce it to Bring-Jerrard form and discussed in this post. A second and rather simpler way is to reduce it to the Brioschi quintic form,

$$w^5-10cw^3+45c^2w-c^2 = 0tag1$$

To solve $(1)$, solve the parameter $m$ as a root of the quadratic $m(1-m) = v$, and where $v$ is a root of the cubic,

$$frac{256(1-v)^3}{v^2}=frac{1728c-1}{c}tag2$$

Then define the argument $tau$ as,

$$tau = ifrac{K(k')}{K(k)}+color{brown}n = i,frac{text{EllipticK[1-m]}}{text{EllipticK[m]}}+color{brown}ntag3$$

with the complete elliptic integral of the first kind $K(k)$ and elliptic parameter $m=k^2$ (with $tau$ also given in Mathematica syntax above). Note that while $(2)$ is a sextic in $m$, it is just a cubic in $v$ and hence is solvable in radicals.

Now that we have $tau$, we can solve $(1)$ in two ways:

Method 1: The Dedekind eta function $eta(tau)$.

The five roots $w_n$ of the Brioschi quintic $w^5-10cw^3+45c^2w-c^2 = 0$ are,

$${w_color{brown}n}^2 =frac{-c,(f^2+4)(f^2-2f-4)^2}{f^5+5f^3+5f-11}tag4$$

where $f:=f_n$ for $color{brown}n = 0,1,2,3,4,$

$$f_n = 1+frac{etabig(tau/5big)}{etabig(5taubig)}tag5$$

Some remarks:

1. Since $(4)$ involves a square, the appropriate sign should be used after taking the square root. (There is another expression without a square root but is more complicated.)


2. The solution implies that the general quintic can be solved by the Rogers-Ramanujan continued fraction,
   $$r(tau)= cfrac{q^{1/5}}{1+cfrac{q}{1+cfrac{q^2}{1+ddots}}}$$
   since if $q = exp(2pi i tau)$, then,
   $$frac{1}{r(tau)}-r(tau) =1+frac{eta(tau/5)}{eta(5tau)}tag6$$
   and one can see the affinity between $(5)$ and $(6)$.

Method 2: The Jacobi theta function $vartheta_2(0,p).;$ (Added Nov 27, 2015.)

The five roots $w_n$ of the Brioschi quintic $w^5-10cw^3+45c^2w-c^2 = 0$ are,

$$w_{n}=pmsqrt{frac{-c,(x^2+4)(x^2-2x-4)^2}{b-11}}$$

with $n=0,1,2,3,4$ where (see also this post),

$$x_n=2sinhBigg(tfrac{sinh^{-1}Big(tfrac{b}{2}Big),+,2pi,i, n}{5}Bigg) = -2isinBigg(tfrac{ilogBig(tfrac{b+sqrt{b^2+4}}{2}Big),-,2pi, n}{5} Bigg)tag7$$

$$b=frac{v(v-5)^2}{(v-1)^2}+11$$

$$v=left(frac{vartheta_2(0,p)}{vartheta_2(0,p^5)}right)^2$$

$$p=e^{pi i tau}=exp(pi i tau)$$

$$tau = ifrac{K(k')}{K(k)} = i,frac{text{EllipticK[1-m]}}{text{EllipticK[m]}}$$

with $m$ (through $v$) as defined by $(2)$. Hence, in addition to continued fractions, step $(7)$ also shows that the general quintic can be solved by trigonometric and hyperbolic functions (though also using special functions).

Note: It also uses the Jacobi theta function $vartheta_j(0,p)$ (where $j=3$ or $4$ will work as well).