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Given some degree 6 polynomial $f(x) \in \mathbb{Q}[x]$, is there any invariant of the polynomial (depending on the coefficents) that will tell you if this polynomial has 6 complex roots or just 2 complex roots.

Edit: By complex root I mean a root in $\mathbb{C}$ but not $\mathbb{R}$.

In other words, is it possible to tell between the two cases without explicitly finding the roots.

Thank you

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    $\begingroup$ You missed this case that the polynomial always have $6$ roots by fundamental theorem of algebra.... $\endgroup$
    – k1.M
    Commented May 14, 2015 at 11:54
  • $\begingroup$ But if you are looking for number of real roots you should use mean value theorem to find the number of real roots...this polynomial has an even number of real roots... $\endgroup$
    – k1.M
    Commented May 14, 2015 at 11:56
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    $\begingroup$ I think I was a big vague with what I meant by complex root. I've edited to question now. $\endgroup$ Commented May 14, 2015 at 12:01
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    $\begingroup$ You can use the discriminant, Descartes' rule of signs, or Sturm's sequences. $\endgroup$
    – lhf
    Commented May 14, 2015 at 12:05
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    $\begingroup$ See also mathoverflow.net/questions/20946/…. $\endgroup$
    – lhf
    Commented May 14, 2015 at 12:09

1 Answer 1

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Yes, there is a closed algebraic expression involving only the coefficients of $f$ such that $f$ has exactly 6 non-real roots if and only if this expression is true thanks to the Tarski–Seidenberg quantifier elimination theorem. In short this says that:

``any system of polynomial equations and inequalities over the reals involving or, and, not, for all and there exists is equivalent to a quantifier free one.''

As having 6 non-real roots is equivalent to having no real roots which is equivalent to the statement: $$ \forall x \in \mathbb{R},\; f(x) \neq 0 $$ we can transform this into a quantifier free statement using the cylindrical decomposition algorithm (this is implemented as Mathematicas Resolve function. However this can quickly become unwieldy, for example when $$ f(x) = x^4 + d x^3 + cx^2 + bx + a $$ the command:

Resolve[ForAll[x, x^4 + d x^3  + c x^2 + b x + a != 0], Reals]

results in:

$$ \left(c<\frac{3 d^2}{8}\land \left(\left(b<\frac{1}{8} \left(4 c d-d^3\right)-\frac{\sqrt{-512 c^3+576 c^2 d^2-216 c d^4+27 d^6}}{24 \sqrt{3}}\land a>\text{Root}\left[256 \text{$\#$1}^3+\text{$\#$1}^2 \left(-192 b d-128 c^2+144 c d^2-27 d^4\right)+\text{$\#$1} \left(144 b^2 c-6 b^2 d^2-80 b c^2 d+18 b c d^3+16 c^4-4 c^3 d^2\right)-27 b^4+18 b^3 c d-4 b^3 d^3-4 b^2 c^3+b^2 c^2 d^2\&,1\right]\right)\lor \left(\frac{1}{8} \left(4 c d-d^3\right)-\frac{\sqrt{-512 c^3+576 c^2 d^2-216 c d^4+27 d^6}}{24 \sqrt{3}}\leq b<\frac{1}{8} \left(4 c d-d^3\right)\land a>\text{Root}\left[256 \text{$\#$1}^3+\text{$\#$1}^2 \left(-192 b d-128 c^2+144 c d^2-27 d^4\right)+\text{$\#$1} \left(144 b^2 c-6 b^2 d^2-80 b c^2 d+18 b c d^3+16 c^4-4 c^3 d^2\right)-27 b^4+18 b^3 c d-4 b^3 d^3-4 b^2 c^3+b^2 c^2 d^2\&,3\right]\right)\lor \left(b=\frac{1}{8} \left(4 c d-d^3\right)\land a>\text{Root}\left[256 \text{$\#$1}^3+\text{$\#$1}^2 \left(-192 b d-128 c^2+144 c d^2-27 d^4\right)+\text{$\#$1} \left(144 b^2 c-6 b^2 d^2-80 b c^2 d+18 b c d^3+16 c^4-4 c^3 d^2\right)-27 b^4+18 b^3 c d-4 b^3 d^3-4 b^2 c^3+b^2 c^2 d^2\&,2\right]\right)\lor \left(\frac{1}{8} \left(4 c d-d^3\right)<b\leq \frac{\sqrt{-512 c^3+576 c^2 d^2-216 c d^4+27 d^6}}{24 \sqrt{3}}+\frac{1}{8} \left(4 c d-d^3\right)\land a>\text{Root}\left[256 \text{$\#$1}^3+\text{$\#$1}^2 \left(-192 b d-128 c^2+144 c d^2-27 d^4\right)+\text{$\#$1} \left(144 b^2 c-6 b^2 d^2-80 b c^2 d+18 b c d^3+16 c^4-4 c^3 d^2\right)-27 b^4+18 b^3 c d-4 b^3 d^3-4 b^2 c^3+b^2 c^2 d^2\&,3\right]\right)\lor \left(b>\frac{\sqrt{-512 c^3+576 c^2 d^2-216 c d^4+27 d^6}}{24 \sqrt{3}}+\frac{1}{8} \left(4 c d-d^3\right)\land a>\text{Root}\left[256 \text{$\#$1}^3+\text{$\#$1}^2 \left(-192 b d-128 c^2+144 c d^2-27 d^4\right)+\text{$\#$1} \left(144 b^2 c-6 b^2 d^2-80 b c^2 d+18 b c d^3+16 c^4-4 c^3 d^2\right)-27 b^4+18 b^3 c d-4 b^3 d^3-4 b^2 c^3+b^2 c^2 d^2\&,1\right]\right)\right)\right)\lor \left(c\geq \frac{3 d^2}{8}\land a>\text{Root}\left[256 \text{$\#$1}^3+\text{$\#$1}^2 \left(-192 b d-128 c^2+144 c d^2-27 d^4\right)+\text{$\#$1} \left(144 b^2 c-6 b^2 d^2-80 b c^2 d+18 b c d^3+16 c^4-4 c^3 d^2\right)-27 b^4+18 b^3 c d-4 b^3 d^3-4 b^2 c^3+b^2 c^2 d^2\&,1\right]\right) $$

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  • $\begingroup$ Wow this is great! Thank you, this is just what I needed. $\endgroup$ Commented May 17, 2015 at 10:41

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