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Let $P(z)$ be an arbitrary polynomial with real coefficients. I'd like to guarantee that all roots of $P$ have real parts outside the interval $(0, 1)$. Is there some simple condition on P that will ensure that?

Let me illustrate further. If $P$ were known to only have real roots, a simple condition would (almost tautologically) be “P does not change sign on $(0, 1)$”. For the general case that $P$ has complex roots, is there some condition that will guarantee that no root $z_i$ satisfies $0 < \mathrm{Re}\, z_i < 1$?

Note that I'm looking for a sufficient condition. That is, I don't need to characterize the class of all polynomials that fulfill the stated condition; a “reasonably” large subclass of that class may be enough for me.

Luis Mendo
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This is a nontrivial task, and the best approach may depend on how the polynomial is specified (i.e., what kind of structure it has). In principle, the Routh–Hurwitz theorem can answer the question, as it gives the number of roots in the positive half plane $\{z:\operatorname{Re}z>0\}$. Your problem amounts to determining whether this number is the same for $P$ and for $P(z+1)$.

The related Routh–Hurwitz stability criterion is easier to check in practice, and provides a sufficient condition: if the criterion holds, there are no roots with positive real part, and in particular with real part between $0$ and $1$.

Finally, since Alex S mentioned Rouché's theorem, I'll point to the post How many roots of a polynomial have positive real part? for an example of its usage in this context.

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  • Thanks! I also thought about Hurwitz polynomials. However, they seem too restrictive, because if I'm correct a Hurwitz polynomial with real coefficients must hace all its coefficients positive. That rules out many Taylor polynomials (truncated Taylor series), which is the application I had in mind. On the other hand, this might be the best one can get. I'll wait for a while in case other answers appear – Luis Mendo Jul 03 '15 at 09:17