How to solve the Brioschi quintic in terms of elliptic functions?

Question

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.

Answer 1

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,

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

To solve $(1)$, first solve for the cubic in $d$,

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

then solve the parameter $m$ as a root of the quadratic $m(1-m) = d$. Then define the argument $\tau$ as,

$$\tau = i\frac{K(k')}{K(k)}+\color{red}n = i\,\frac{\text{EllipticK[1-m]}}{\text{EllipticK[m]}}+\color{red}n\tag3$$

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). 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,

$${w_n}^2 =\frac{-c\,(f^2+4)(f^2-2f-4)^2}{f^5+5f^3+5f-11}\tag4$$

where for $\color{red}n = 0,1,2,3,4,$

$$f_n = 1+\frac{\eta\big(\tau/5\big)}{\eta\big(5\tau\big)}\tag5$$

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 $q = \exp(2\pi i \tau)$, then, $$\frac{1}{r(\tau)}-r(\tau) =1+\frac{\eta(\tau/5)}{\eta(5\tau)}\tag6$$ 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,

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

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

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

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

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

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

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

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).