I have the polynomial


where the variable $a$ is a random variable from the uniform distribution in the range $[0,1)$. When I analyze this function using Mathematica and specific values for $a$, I find there are four complex roots, a negative real root, and a positive real root - I need to extract the positive real root to get an equation $x=x^+(a)$.

I have tried both of Kulkarni's methods${}^{1,2}$ in which he suggests splitting the sextic equation into a quadratic and a quartic or a cubic and another cubic, but numerics in Mathematica show this cannot be done.

At this point, I'm leaning toward using the Kampe de Feriet functions - these are referenced in many places but never given explicitly.

Could someone provide a hint to a solution, or a reference where I could find the application of the Kampe de Feriet function to sextic equations?

${}^1$http://elib.mi.sanu.ac.rs/files/journals/tm/21/tm1124.pdf ${}^2$http://euclid.trentu.ca/aejm/V3N1/Kulkarni.V3N1.pdf

Here's what the plot looks like near the region I'm interested in, for $a\in[0.1,0.9]$:


  • 1
    $\begingroup$ The only helpful thing I can see is that the polynomial can be written as $3a(x^{2}-1)^{3}+6x^{5}-4x^{3}+6x$. I can't see how to take this any further, though (and I'm sure you'd have noticed if one of your roots was always $\pm1$). $\endgroup$ – preferred_anon Jan 14 '16 at 22:34
  • $\begingroup$ @DanielLittlewood: His sextic, in fact, had a solvable Galois group. I also made an answer. $\endgroup$ – Tito Piezas III Mar 23 '16 at 21:14

Actually, your sextic has a solvable Galois group so can be exactly expressed in terms of radicals. The solution turns out to be quite simple. Given

$$3a x ^6 + 6x^5 - 9a x^4 - 4x^3 + 9a x^2 + 6x - 3a = 0\tag1$$

Then the two real roots are,

$$x = u\pm\sqrt{u^2+1}\tag2$$



and $z_1$ is any non-zero root of,


Your sextic's positive real root $x$ is the positive case $\pm$ of $(2)$.

Note: Be careful when testing this numerically, especially using $(3)$. What I find supremely annoying about my version of Mathematica is it gives a complex value to $z^{1/3}$ when $z$ is a negative real number.

P.S.: However, an easy way to verify this is to use resultants and the Mathematica syntax,

Factor[Resultant[x^2 - 2u x - 1, a u^3 + u^2 + 1/3, u]]

which then yields $(1)$, showing that your sextic in just a quadratic $(2)$ in $x$, with coefficients that is cubic $(3)$ in $u$.


After some thought I've decided to use a root-finding method (Newton's method) since I'm working with a polynomial. It can be factored into the form

$$f(x;a) = -3a + x(6 + x(9a + x(-4 + x(-9a + x(6 + 3ax)))))$$

which is most efficient computationally. Ideally I'd also like to know the theoretical expression, but it's unlikely it would be faster to compute than a simple polynomial.

Again, if anyone can point me to literature explaining Kampe de Feriet functions and their applications to polynomial roots, I'd be thankful.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.