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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

  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, $z^6+z^2+3z+2=0$?
  • Related questions about the Kampé de Fériet function here and here. – Tito Piezas III Jun 30 '23 at 15:12
  • couldn't you just write $z=\beta q/\alpha$, multiply through by $1/\beta$, and then you would be left with $$\frac{\beta^5}{\alpha^6}q^6+\frac{\beta}{\alpha^2}q^2+q+1=0$$ – clathratus Jun 30 '23 at 17:52
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    @clathratus Of course, that’s why i said it is just the general sextic in disguise. The problem is R. Israel’s soluton has a narrow radius of convergence. There has to be a way to extend it – Tito Piezas III Jun 30 '23 at 17:59
  • This article solves the sextic using theta functions defined as a double sum. – Тyma Gaidash Jul 06 '23 at 12:39
  • @TymaGaidash Thanks for the link. Considering the intricacies of the procedure, the author should have included a single numerical example as added proof. Do you know Niels Abel actually thought he found the quintic formula in radicals? Two math professors could not find the flaw, until Degen asked for a numerical example. It was then Abel found the error, reversed direction, and proved it was impossible to solve. Just a thought. :) – Tito Piezas III Jul 06 '23 at 13:01

1 Answers1

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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 $|a|,|b|>1$.

Тyma Gaidash
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    The sums look like an Appell hypergeometric series, but it is hard to see how to convert the factorials/gamma functions of $\frac n6$ into pochhammer symbols $(a)_n$ like in the link. At least we have a larger convergence region. – Тyma Gaidash Jun 30 '23 at 20:54
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    If one really wanted to, there is easily a “closed form” with a so called Srivastava Daoust function – Тyma Gaidash Jun 30 '23 at 23:43
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    Nice! This seems faster since it is not a double-summation, has broader convergence, though still not the complete solution to the reduced sextic. (P.S. I made some minor changes to the infobox, and added the form using the hypergeometric function ${_2F_1}$. – Tito Piezas III Jul 01 '23 at 06:11
  • I have an important question. Will your analysis still work if we go higher to the reduced septic, or $x^7+(a+x)(b+x)(c+x) = 0$? Will a hypergeometric function still solve it, or no more? If it can, i will ask a separate question. – Tito Piezas III Jul 02 '23 at 01:03
  • @TitoPiezasIII Yes, as a triple hypergeometric series because the Lagrange reversion coefficients would be $\left.\frac{d^{n-1}}{dx^{n-1}}(x+a)^\frac17(x+b)^\frac16(x+c)^\frac17\right|_{x=0}$. Applying general Leibniz rule twice, or multinomial product rule formula from the link, will get you a double sum for the series coefficients. – Тyma Gaidash Jul 02 '23 at 01:08
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    I have made the new question. Kindly check this post. – Tito Piezas III Jul 02 '23 at 01:56
  • @TitoPiezasIII If it is of interest, the Mathematica function inverse beta regularized can solve $f(x)=cx^6+b x^2+a x+\frac{25 a^2}{96 b}=0$ for some $x$. If one expands the regularized beta function in the “solve” link, it would be equivalent to solving $f(x)$. However, one more free parameter is needed for solving $g(x)=x^6+x^2+ax+b=0$. Maybe $f(x)$ can transform into $g(x)$. – Тyma Gaidash Jul 05 '23 at 02:15
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    We can set $c=1$ without loss of generality. The problem is the constant term is a function of the $x^2,x$ terms. Scaling variables will not do it. And it already took a quartic Tschirnhausen to eliminate 3 terms; to satisfy this additional constraint may require a higher one, so I'm afraid we're stuck. Your sextic does have the nice form $$96b x^6 + 24(2b x + a)^2 + a^2 = 0$$ and its Galois group has order 6!=720, the highest for deg 6. – Tito Piezas III Jul 05 '23 at 06:25
  • To all: there is probably a way to find all complex solutions in the form $z=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}x^{-s}\text{Gauss hypergeometric function expression}ds$ or, expanding the integrand, as a double integral. This solution can be written if some users wish – Тyma Gaidash Jul 23 '23 at 12:20
  • Kindly go ahead. It is allowed to give two answers if the other gives another perspective. – Tito Piezas III Jul 23 '23 at 19:33
  • If the “two parameter fox h function” were standardized, then we likely would have a closed form using the double integral which may be attempted at another time. – Тyma Gaidash Nov 05 '23 at 14:11
  • Even though it is not the Kampé de Fériet function, it is still a very useful answer that applies to degrees $5,6,7,8$. I should have clicked on the button a long time ago. Mea culpa. – Tito Piezas III Dec 08 '23 at 16:38