4
$\begingroup$

I'm working with the expression $m = 32an^6+96an^5+120an^4+80an^3+28an^2+4bn^2+4an+4bn+2a+2c $. What exactly is the closed radical form of this, if one were to write $n$ in terms of $m$? There's no general formula for sixth powers but I want to know if this particular class can be given in radicals for all $a, b \in \mathbb{R}$.

$\endgroup$
1
  • 4
    $\begingroup$ Erm, when is that sextic function one-to-one? $\endgroup$ Jun 23 at 4:23

2 Answers 2

10
$\begingroup$

The RHS only depends on $n(n+1)$, specifically it can be written as $$ 4a\left(8(n(n+1))^3+6(n(n+1))^2+n(n+1)+\frac12\right)+4bn(n+1)+2c, $$ so you have to use the standard formulas to solve a cubic equation $g(n(n+1))=m$, and then a quadratic equation $n(n+1)=k$, which can be done in radicals.

$\endgroup$
1
  • $\begingroup$ neat; any idea how a CAS such as Maple or Mathematica finds this decomposition? $\endgroup$ Jun 24 at 7:25
6
$\begingroup$

Mathematica finds a closed-form expression for the solutions $n$ as a function of $m$ of the equation $$m = 32an^6+96an^5+120an^4+80an^3+28an^2+4bn^2+4an+4bn+2a+2c.$$ The expressions for general $a,b,c$ are lengthy. By way of example, for $a=1$, $b=2$, $c=-3$ the two real solutions are $$n=-\tfrac{1}{2}\pm\tfrac{1}{2}\sqrt{f_m^{1/3}-f_m^{-1/3}},\;\;f_m=\sqrt{m^2+12 m+37}+m+6.$$

More generally, two of the solutions are $$n=-\frac{1}{2}\pm\frac{1}{8}\sqrt{\frac{2^{2/3} \left(\sqrt{4 s^3+z^2}+z\right)^{2/3}- 2^{4/3} s}{3a \left(\sqrt{4 s^3+z^2}+z\right)^{1/3}}},$$ $$z=27648 a^2 (b-2a)+27648 a^2 (m-2c),\;\;s=192 a(2b-a).$$ I have not checked the parameter range where these two solutions are real.

$\endgroup$
2
  • $\begingroup$ Well I'll be... $\endgroup$ Jun 23 at 11:53
  • 5
    $\begingroup$ Surely this indicates that there is something special about the OP's sextic... $\endgroup$ Jun 23 at 15:00

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.