First, some background.
I. One-Parameter forms
The Bring-Jerrard quintic, $x^5+x+\alpha=0,$ has a solution as,
$$x = -\alpha\sum_{k=0}^\infty(-1)^k\frac{(5k)!}{k!(4k+1)!}\;\alpha^{4k}$$
This series has a narrow radius of convergence, namely $|\alpha|<\left(\frac{4^4}{5^5}\right)^{1/4}\approx 0.53$. But it can be extended via analytic continuation using the generalized hypergeometric function ${_pF_q},$
$$x = -\alpha\,{_4F_3}\Big(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,-\frac{5^5}{4^4}\alpha^4\Big)$$
for more general $\alpha$, thus solving the general quintic.
II. Two-Parameter forms
The quintic and sextic two-parameter equations,
$$By^5+Ay^2+y+1 = 0$$ $$Bz^6+Az^2+z+1 = 0$$
can be solved as,
$$y = -\sum_{j=0}^\infty \sum_{k=0}^\infty (-1)^k \frac{(2j+5k)!}{j!k!(j+4k+1)!}\;A^j B^k$$
$$z = -\sum_{j=0}^\infty \sum_{k=0}^\infty (+1)^k \frac{(2j+6k)!}{j!k! (j + 5 k + 1)!}\; A^j B^k$$
with the quintic root $y$ by Passare and Tsikh in this paper, and the sextic root $z$ by Robert Israel in this old MSE answer. The second equation is just the general sextic in disguise since the general sextic can be reduced to the form,
$$z^6+z^2+\alpha z+\beta =0$$
Unfortunately, R. Israel's solution $z$ also has a narrow radius of convergence. So we are seeking an analytic continuation such that it will be valid for more general $(A,B)$ thus solving the general sextic.
III. Questions
- If the analytic continuation for the quintic $x$ involves the one-parameter generalized hypergeometric function, does the analytic continuation for R. Israel's sextic $z$ involve the two-parameter Kampé de Fériet function?
- Or is it some other two-parameter function, maybe like the Appell series, Humbert series, etc?
- The Kampé de Fériet function can solve the general sextic in its reduced form. In Mathematica syntax, what would be the input to solve, for example, $z^6+z^2+3z+2=0$?