This is the third in my series of problems for enthusiasts, a full list of which can now be found on my profile! Apologies, as usual, if you find them insultingly trivial.
A complex polynomial is called Hurwitz if each one of its roots has a negative real part. Such polynomials are of central importance in linear control theory. If $p$ an $q$ are two polynomials, we denote their composition by $p\circ q$, which is defined as the polynomial $r$ such that $r(x) = p(q(x))$ for all complex $x$. Our first question is immediate:
- When is the composition of two Hurwitz polynomials itself Hurwitz?
Suppose $r$ is Hurwitz and admits the representation $r = p_1\circ p_2\circ \dots \circ p_n$ ($n \geq 2$) for Hurwitz polynomials $p_i$ of degree strictly less than $r$. In this case we say $r$ is Hurwitz factorizable and call this representation the Hurwitz factorization of $r$ with factors $p_i$. The set is nonempty--in fact, it's uncountably infinite--for it contains $x^4 +2ax^2 + a^2 +a = (x^2+a)\circ(x^2+a)$ for all $a > 0$. If a Hurwitz polynomial does not possess a Hurwitz factorization, it is called irreducible.
Furthermore, if one or more factors of a Hurwitz factorization are themselves Hurwitz factorizable, another representation of the first factorization is possible using the factors of the second. If this process of factoring is continued, the degree condition implies it must terminate after a finite number of steps (if the degree condition is not enforced it need not! Consider $x+a = (x+a/2)\circ (x+a/2)$.) Thus this process ends exactly when every factor is irreducible. This invites question
Given an arbitrary Hurwitz polynomial,
(a) determine whether or not it is irreducible and, if not,
(b) provide an algorithm to calculate its irreducible Hurwitz factorization.