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Descartes' rule of signs claims that there exist an upper bound on the number of real roots of a polynomial based on the sign differences between consecutive nonzero coefficients.

But, there exists some similar criterion to obtain a lower bound on the number of real roots?

P.S. It's clear that for polynomials with odd degree, a real root is guaranteed.

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  • $\begingroup$ By the same reasoning, for polynomials of even degree, the lower bound on the number of real roots is $0$. $\endgroup$ – Andy Walls Apr 24 '18 at 12:53
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As @Andy Walls mentioned in his comment, an even-degree polynomial with real coefficients is not guaranteed any real roots - $y=x^2+1$ is a simple example. If the polynomial is of odd degree and has real coefficients, you can prove by the Intermediate Value Theorem that it has at least one real root, call it $a$. But then, if you factor out the expression $x-a$ from the polynomial, what's left is of even degree, and itself is not guaranteed any real roots.

If you have a polynomial of even degree with real coefficients, and an odd number of sign changes in those coefficients, then you're guaranteed two real roots, because complex roots always come in complex conjugate pairs.

Summary: odds have minimum one root, evens have minimum zero roots unless there are an odd number of sign changes, in which case you're guaranteed two real roots.

All bets are off if you have complex coefficients.

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  • $\begingroup$ But my question is about the relationship between the coefficients and signs' coefficients of a polynomial and its roots, not for a general one. I.e., is there some criterion that ensures us the existence of a real root in an even degree polynomial? For example, we can ensure the existence of real roots of the polynomial $x^4 - 5x^2 + 5$ without calculation via some criterion similar to Descartes' signs? $\endgroup$ – ted711 Apr 24 '18 at 17:57
  • $\begingroup$ I don't know of any criterion such as you suggest. There simply is no lower bound (except zero) on the number of real roots of a polynomial of even degree, except for Descartes' Rule of Signs. If there's only one sign change in the coefficients of a polynomial of even degree, then I suppose that would force you to have two real roots, because complex roots always come in complex conjugate pairs. But if an even-degreed polynomial has an even number of sign changes, you could easily have zero real roots. $\endgroup$ – Adrian Keister Apr 25 '18 at 13:23

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