ca.classical-analysis-and-odesinequalitiespolynomials
Suppose we have a polynomial function $f(z)=a_0+a_1z+a_2z^2+…+z^n$ with each $a_i$ between 0 and 1. Is there a method to determine if $f$ has a pair of complex conjugate roots?
There are many results on radius of roots, but I never see similar facts concerning the question here.
There is indeed an easy way to check if a univariate poly with real coefficients has a real root, without computing the roots.
Note that the answer for odd degree polynomials is always yes. For an even degree polynomial $p(x)$ do the following:
Compute the Hermite form of the polynomial. This is a symmetric matrix defined e.g. here: (on page 4, near the bottom denoted by $H_1(p)$). The entries of this matrix can be filled using Newton identities.
The number of real roots of $p(x)$ is equal to the signature of the Hermite matrix $H_1(p)$, i.e., the number of positive eigenvalues minus the number of negative eigenvalues.
Since the Hermite form is symmetric, its characteristic poly has only real roots. Therefore, we can apply Descartes' rule of signs to its char. polynomial, which would give us exactly the number of positive and negative eigenvalues of the Hermite form.
This process gives you the exact number of real roots of $p(x)$ without computing the roots, and in particular you can see if the polynomial has a real root.
Hope this helps.
-Amirali Ahmadi
First, a counterexample to your conjecture. Let $\Pi = x^4+x^3+4x^2+4x = x(x+1)(x^2+4)$, so $P = 4x^3+3x^2+8x+4$. The critical values of $\omega$ are $1.06638, 3.89455 + 2.87687i, 3.89455 - 2.87687i$, and by inspection (using Mathematica) we see that for each of these values of $\omega$, $\mbox{Conv}(\Pi_\omega)$ contains a neighborhood of $0$.
Now for a calculus on convex sets. Every convex set is the intersection of a set of halfplanes. Call a halfplane in this collection essential if removing all of the halfplanes in an open set of halfplanes (in the halfplane topology) containing it from our set of halfplanes makes the intersection of the halfplanes in our set bigger. We wish to find a characterization of the essential halfplanes of $\mbox{Hull}(P)$.
First of all, I claim that any essential halfplane of $\mbox{Hull}(P)$ occurs as an essential halfplane of $\mbox{Conv}(\Pi_\omega)$ for some $\omega$. This follows from continuity - for any open set around our essential halfplane there is some $\omega$, take the limit of a subsequence of these $\omega$s...
Now, suppose that the halfplane $\mbox{Re}(x) \le 0$ occurs as an essential halfplane of some $\mbox{Conv}(\Pi_\omega)$, i.e. there are at least two roots of $\Pi_\omega$ with real part $0$, and the rest of the roots have negative real part. If the number of roots on the line $\mbox{Re}(x) = 0$ (counted with multiplicity) is two, then by holomorphicity we can always find a direction to move $\omega$ so that either both roots move to the left, or both roots stay on the line $\mbox{Re}(x) = 0$ and move towards eachother. If we can ever make both roots move to the left, then clearly the halfplane $\mbox{Re}(x) \le 0$ is not an essential halfplane of $\mbox{Hull}(P)$, otherwise we keep pushing the roots towards eachother until either they run into eachother or until a third root hits the line $\mbox{Re}(x) = 0$. In any case, we see that if a halfplane is essential for $\mbox{Hull}(P)$, then there is some $\omega$ such that the halfplane is essential for $\mbox{Conv}(\Pi_\omega)$ and such that at least three roots (counted with multiplicity) of $\Pi_\omega$ are on the boundary of the halfplane, or two of the roots are equal and $\Pi_\omega$ has no other roots.
So if we let $L$ be the set of $\omega$s such that three of the roots of $\Pi_\omega$ lie on a line, we get that $\mbox{Hull}(P) = \cap_{\omega \in L} \mbox{Conv}(\Pi_\omega)$ if $\deg P \ge 2$.
Edit: Actually, I think there is a problem with this. It's conceivable that two roots are on the line $\mbox{Re}(x) = 0$ and have derivatives (with respect to $\omega$) pointed in opposite directions, such that we can't simply push them towards eachother. For instance, the map from one root to the other root could, locally, look like the fractional linear transform sending the left halfplane to a circle contained in the right halfplane and tangent to the line $\mbox{Re}(x) = 0$ at the other root. So, we may need to enlarge the set $L$ to contain also those $\omega$s for which the ratio of the derivatives of two of the roots (with respect to $\omega$) is a negative real number.
Edit 2: It turns out that this isn't an issue. Call the two roots on the line $\mbox{Re}(x) = 0$ $r_1$ and $r_2$. Suppose that locally, $r_1(\epsilon) = \epsilon$, $r_2(\epsilon) = i - m\epsilon + a\epsilon^k + O(\epsilon^{k+1})$, $a \ne 0$, $m > 0$. Note that if we had $r_2(\epsilon) = i-m\epsilon$, then the intersection of the halfplanes corresponding to $r_1(\epsilon), r_2(\epsilon)$ and $r_1(-\epsilon), r_2(-\epsilon)$ would be contained in the halfplane $\mbox{Re}(x) \le 0$, and the intersection of their boundaries would be located at $i/(m+1)$. Now if $k$ is even, then the correction term shifts the intersection of the boundaries by $a\epsilon^k/(m+1) + O(\epsilon^{k+1})$, so if we choose $\epsilon$ small such that $a\epsilon^k$ is real and negative, then we see that the halfplane $\mbox{Re}(x) \le 0$ is not essential. If $k$ is odd, then if we choose $\epsilon$ small such that $\mbox{Re}(\epsilon) < 0$ and $a\epsilon^k$ is a positive real times $i$, then $r_2(\epsilon)$ is shifted up and $r_2(-\epsilon)$ is shifted down, so the intersection of the boundaries will be shifted to the left (draw a picture), so again the halfplane $\mbox{Re}(x) \le 0$ is not essential.
Best Answer
We can assume $f$ has no multiple root (if the gcd of $f$ and $f'$ is not constant, divide by this gcd). Let $n$ be the degree of $f$. Compute $$\frac{f(X)f'(Y)-f(Y)f'(X)}{X-Y} = \sum_{i,j=i}^{n}a_{i,j}\; X^{i-1}\; Y^{j-1}\;.$$ Then $f$ has all roots real iff the symmetric matrix $(a_{i,j})_{i,j=1,\ldots,n}$ is positive definite. This can be checked for instance by computing the principal minors of this matrix and verifying whether they are all positive.
There are several methods for computing the number of real roots using signature of quadratic form : see for instance this note (in french).