The trigonometric solution to the solvable DeMoivre quintic?

Question

Using the relations for the Rogers-Ramanujan cfrac described in this post,

$$\frac{1}{r}-r = x$$

$$\frac{1}{r^5}-r^5 = y$$

and eliminating $r$ yields,

$$x^5+5x^3+5x = y$$

This is the case $a=1$ of the solvable DeMoivre quintic,

$$x^5+5ax^3+5a^2x+b = 0\tag1$$

In general, it has the solution,

$$x = \left(\tfrac{-b+\sqrt{b^2+4a^5}}{2}\right)^{1/5}-a\left(\tfrac{-b+\sqrt{b^2+4a^5}}{2}\right)^{-1/5}\tag2$$

When $a=-1$, it can also be solved as,

$$x = 2\cos\Bigg(\tfrac{\arccos\Big(\tfrac{-b}{2}\Big)}{5}\Bigg) = 2\cos\Bigg(\tfrac{2\arctan\Big(\tfrac{\sqrt{2+b}}{\sqrt{2-b}}\Big)}{5}\Bigg)\tag3$$

Q: When $\color{red}{a=1}$, is there a neat trigonometric solution similar to $(3)$?

P.S. A google search revealed that it is may indeed be possible.

Answer 1

Thanks to Tito for the nice question. Here is a solution in terms of hyperbolic sine, which may not be what you want. $$ \sinh(5t) = 5 \sinh t+ 20 \sinh^3 t + 16 \sinh^5 t. $$ With $x = 2\sinh t$ and $b = -2\sinh 5t$, we have $$ x^5+ 5 x^3 + 5 x + b = 0\tag1, $$ which is the $a = 1$ case. So the solution is $$ x = 2\sinh t = 2\sinh\left( \frac{\sinh^{-1}\left(-\frac{b}{2}\right)}{5} \right)\tag2 $$

Answer 2

(This is an addendum to hbp's answer.)

I'm glad I raised a bounty for this question because, thanks to hbp, we can show that the general quintic can be solved in terms of trigonometric or hyperbolic functions, plus some special functions.

The general quintic can be reduced to the one-parameter Brioschi form,

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

with the five solutions (see also this post),

$$w_n=\pm\sqrt{\frac{-c\,(x^2+4)(x^2-2x-4)^2}{b-11}}\tag2$$

and for $n=0,1,2,3,4$,

$$x_n=-2\,i\sin\Bigg(\tfrac{i\log\Big(\tfrac{b+\sqrt{b^2+4}}{2}\Big)\,-\,2\pi\, n}{5} \Bigg)=2\sinh\Bigg(\tfrac{\sinh^{-1}\Big(\tfrac{b}{2}\Big)\,+\,2\pi\,i\, n}{5}\Bigg) \tag3$$

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

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

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

$$\tau = i\frac{K(k')}{K(k)}= i\,\frac{\text{EllipticK[1-m]}}{\text{EllipticK[m]}}\tag4$$

$$m =\tfrac{1}{2}\left(1\pm\sqrt{1-4u}\right)\tag5$$

and $u$ is a root of the cubic,

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

The solution also uses the Jacobi theta function $\vartheta_j(0,p)$ (where $j=3$ or $4$ will work as well), 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 1: Since $(2)$ and $(5)$ uses square roots, then the proper sign has to be chosen.

Note 2: At first, I was missing a neat form for $(3)$ so, after hbp's answer, I'm glad I posted this question.