Solving the reduced sextic $z^6+z^2+az+b=0$ using the two-parameter Kampé de Fériet function?

Question

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.

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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.

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III. Questions

1. 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?
2. Or is it some other two-parameter function, maybe like the Appell series, Humbert series, etc?
3. 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, $z^6+z^2+3z+2=0$?

Answer 1

The asker wanted an analytic continuation, so we apply @Richard Stanley’s method in “Series solution for general trinomial” with Lagrange reversion:

$$\begin{align}z^6+z^2+az+b=0\implies z_k=e^\frac{(2k+1)\pi i }6(z^2+az+b)^\frac16=\omega_k(z-p)^\frac16(z-q)^\frac16=\sum_{n=1}^\infty\frac{\omega_k^n}{n!}\left.\frac{d^{n-1}}{dz^{n-1}}(z-p)^\frac n6(z-q)^\frac n6\right|_0,k=0,\dots ,6\end{align}$$

General Leibniz rule uses factorial power $n^{(m)}$:

$$\left.\frac{d^{n-1}}{dz^{n-1}}(z-p)^\frac n6(z-q)^\frac n6\right|_0=\sum_{m=0}^{n-1}\binom{n-1}m\left(\frac n6\right)^{(n-m-1)}\left(\frac n6\right)^{(m)}(-p)^{m-\frac{5n}6+1}(-q)^{\frac n6-m}$$

Summing over $m$ presents a regularized Gauss hypergeometric function ${_2\tilde{\text F}_1}$:

$$\bbox[2.5px,border:5px groove blue]{\begin{align} &z^6+z^2+az+b=0\implies \\ z_k &=\sum_{n=1}^\infty\sum_{m=0}^{n-1}e^\frac{(2k+1)\pi i n}6\frac{\left(\frac n6\right)!^2 (-p)^{m-\frac{5n}6+1}(-q)^{\frac n6-m}}{\left(m-\frac{5n}6+1\right)!\left(\frac n6-m\right)!\Gamma(n-m)m!n}\\ &=\sum_{n=1}^\infty\frac{\left(\frac n6\right)!}{n!}\,e^\frac{(2k+1)\pi i n}6(-p)^{1-\frac{5n}6}(-q)^\frac n6\,_2\tilde{\text F}_1\left(1-n,-\frac n6;2-\frac{5n}6;\frac pq\right)\\ \text{where},\\ p,q&=\frac{-a\pm\sqrt{a^2-4b}}2\end{align}}$$

It is interesting to note the $6n$ terms are $0$. Alternatively, using the hypergeometric function ${_2 F_1}$,

$$z_k=\sum_{n=1}^\infty\frac{\left(\frac n6\right)!}{n!\,\Gamma\big(2-\frac{5n}6\big)}\,e^\frac{(2k+1)\pi i n}6(-p)^{1-\frac{5n}6}(-q)^\frac n6\,{_2 F_1}\left(1-n,-\frac n6;2-\frac{5n}6;\frac pq\right)$$

The example $z^6+z^2+3z+2=z^6+(z+2)(z+1)=0$ cannot be solved with Robert Israel’s series, as $\frac{32}{729}w^6+\frac29 w^2+w+1=0$ where $w=\frac32 z$ is outside the series convergence. However, our series gives all $6$ roots matching the actual roots. The series usually converge for about $|a|,|b|>1$.