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

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,

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

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).

• @Nicco: Now you know why I was looking for how to express $r(\tau)$ in terms of the Jacobi theta functions. :) – Tito Piezas III Sep 7 '15 at 14:52
• @ Tito Piezas:Yes I do.How about putting some more detail into your answer about method $2$ – Nicco Sep 7 '15 at 15:01
• I've read Duke's paper before, but I'll have to read it again. Then I'll have to see if the resulting expressions are as simple as for Method 1. – Tito Piezas III Sep 7 '15 at 15:05
• @Nicco: I finally found a rather simple formulation using the Jacobi theta functions and hyperbolic functions. See Method 2. – Tito Piezas III Nov 27 '15 at 7:36
• @ Tito PiezasIII :What a beautiful method.The fact that there's a connection between hyperbolic functions and elliptic functions is very interesting. – Nicco Mar 28 '16 at 17:06