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For a study I'm making about the minimum and maximum radial values of bounded orbits in a central force system with general power law forces, I came across this special polynomial equation:

$${x^\alpha } + {x^{\alpha - 1}} + \ldots + {x^3} + {x^2} -p= 0, \qquad p>0 ,$$

that is

$\sum\limits_{k = 0}^{\alpha - 2} {{x^{\alpha - k}}} -p=0 .$

The polynomial equation is special since:

  • all the coefficient of the terms with degree $ \ge 2$ are equal
  • the coefficient of the term with degree $1$ is $0$
  • the coefficient of the term with degree $0$ is $-p$ with $p>0$

Using Descartes’ rules of sign to count the number of real positive zeros of above equation (and seeing the numerical solutions given by Mathematica for positive values of $p$) I know that above equation has a single real positive solution.

I'm also aware that there aren't, in general, closed algebraic forms for the solutions of polynomial equations with degree $ \ge 5$ but I wonder if, given the special form of the equation, it is possible to express the single real positive root in closed form.

Clearly the algebraic expression for the real positive root would depend on $p$ and $\alpha$.

Even if there's no closed form for the real positive root maybe there are tools for exploring above special equation (I'm no expert in this field) that I don't know of.

Any link/suggestion is much appreciated.

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  • $\begingroup$ You don't need all that work to see that there is a single positive root - it is clearly increasing on the positive reals and negative at $x=0$, so it has one positive real root. $\endgroup$ Feb 10, 2016 at 14:54
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    $\begingroup$ Although there is no closed formula for $\alpha \gt 4$, the usual numerical root finding algorithms will have no difficulty giving as much accuracy for the unique positive root as you wish. $\endgroup$
    – hardmath
    Feb 10, 2016 at 15:00
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    $\begingroup$ Degree is a much more common term for what in the question is called a grade. $\endgroup$ Feb 10, 2016 at 15:00
  • $\begingroup$ Ok made correction grade->degree $\endgroup$
    – Luca M
    Feb 10, 2016 at 15:09
  • $\begingroup$ You are solving, $$\frac{x^n-1}{x-1} = x+p+1$$ or equivalently, $$(x^n-1)-(x-1)(x+p+1) = x^n-x^2-px+p = 0\tag1$$ Since you didn't specify what kind of closed-form you want, I think a quadrinomial like $(1)$ can be solved in terms of the hypergeometric function, though I am not sure. For comparison on the trinomial, see eq.42 of this article. $\endgroup$ Feb 10, 2016 at 15:30

1 Answer 1

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In general, there is no algebraic formula for such a root: In the case $\alpha = 5$, $p = 1$, for example, the Galois group of the resulting polynomial, $$x^5 + x^4 + x^3 + x^2 - 1 ,$$ is the full symmetric group $S_5$. Since this group is not solvable, there is no expression for any of its roots in terms of basic operations and root extraction.

Of course, in some special cases one may still be able to find a formula; for example, when $p = \alpha - 1$, the unique positive solution is $x = 1$.

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  • $\begingroup$ That's what I suspected. Much useful answer. Could you give an example of a 5th degree polynomial that would be solvable? $\endgroup$
    – Luca M
    Feb 10, 2016 at 15:04
  • $\begingroup$ E.g. $x^5+x^4+x^3+x^2-4=0$ having a root $x=1$, and so it reduces to a fourth degree polynomial, so it has 5 closed-form roots $\endgroup$
    – user304329
    Feb 10, 2016 at 15:08
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    $\begingroup$ Sure, and one can use the special case I mention to generate such an example: $q(x) = x^5 + x^4 + x^3 + x^2 - 4$ has root $1$, so it factors as $(x - 1) r(x)$ for some quartic $r$. Thus, the the Galois group of $q$ is isomorphic to the Galois group of $r$, and so is contained in $S_4$, which is solvable, and in particular there is an explicit algebraic formula for its roots. (In fact, $\operatorname{Gal}(q)$ turns out to be $D_8$.) $\endgroup$ Feb 10, 2016 at 15:09
  • $\begingroup$ @LucaM The following 5th degree polynomial has easily found solutions: $$0=x^5-1$$ $\endgroup$ Feb 10, 2016 at 21:36

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