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Let $p(x)$ be an arbitrary polynomial with real coefficients. Is there a convenient way to determine whether $p(x)$ has a unique real root (not counting multiplicity)?

Edit: I know about Strum's Theorem, but I was hoping that in this special case, there would be a simpler way than computing the entire sequence of Strum polynomials.

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    $\begingroup$ You can start with Descartes Rule of Signs: en.wikipedia.org/wiki/Descartes%27_rule_of_signs $\endgroup$ – Ethan Bolker Nov 30 '17 at 15:42
  • $\begingroup$ For polynomials of even degree, a unique real root would have to be of even multiplicity (and hence necessarily rational, if the coefficients are all rational, so you'll be able to find it). $\endgroup$ – Barry Cipra Nov 30 '17 at 15:51
  • $\begingroup$ Do you know Galois Theory? $\endgroup$ – Stella Biderman Nov 30 '17 at 16:20
  • $\begingroup$ @ Stella Biderman: I have some familiarity with it $\endgroup$ – Mike Hawk Nov 30 '17 at 19:41
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One obviously sufficient answer is: if it derivate is always nonnegative or nonposititive, then is must have unique real root. But this is not necessary to have unique root.

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Here's the best I came up with:

If $p$ has odd degree, it has a real root $a$ so divide through by $x-a$ to the power of its multiplicity and continue as in the even case to verify there are $0$ real roots left.

Let $a_n$ be the leading coefficient of $p$. If $p$ has even degree, find the real roots $r_1,...r_k$ of $p'(x)$ and consider $p(r_1),...p(r_k)$. If they all have the same sign as $a_n$ there are no real roots, if one is $0$ and all others have the same sign as $a_n$ there is exactly one real root, otherwise there's more than one real root.

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