Roots of polynomial with positive coefficients

Question

My question is very simple. Suppose we have a polynomial defined as follows: $$p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots+a_0 $$ where all of the $a_n$'s are all real and positive. Is there something that we can say about the roots of $p(x)$? Can we say the roots of $p(x)$ all contain negative real parts?

Thanks!

Answer 1

You certainly can not say that the real part of any root is negative. Consider $$x^3 +24\sqrt3 = 0.$$ This has ($\sqrt3 + 3i)$ as a root.

Answer 2

The answer is yes, if and only if you are interested in real roots. You can verify it by using Descarte's rule of signs. But in the field of complex numbers, $x^2+1=0$ is the simplest one for contradiction.

ERRATUM (April 18,2021)- Construct a polynomial with roots $-1, \pm i$ to be served as a counterexample i.e. $(x+1)(x-i)(x+i)$ or $x^3+x^2+x+1$

Answer 3

The simplest counter-example is $\;x^5+x^4+x^3+x^2+x+1$: its roots are $$\mathrm e^{\tfrac{\mathrm ik\pi}3},\quad k=1,\dots,5$$ and two of them have positive real parts.

Of course, this polynomial is not irreducible in $\mathbf Q[x]$.

Answer 4

Can we say the roots of p(x) all contain negative real parts?

As pointed out in another answer, this is not true (in general).

However, there is an exlusion sector for such polynomials. Clearly, such polynomials can not have positive roots; however, Aaron Melman recently re-discovered the following result:

Theorem: If $p(t) = \sum_{k=0}^n a_{k}t^{k}$ has nonnegative coefficients and $a_n a_0 \ne 0$, then $p$ has no zeros in the sector $$ S_n = \left\{ z \in \mathbb{C} : \vert \text{Arg}~z \vert \le \pi/n \right\}. $$

Thus, all degree-two polynomials with nonnegative coefficients must have zeros in the left-half plane.

For two elementary proofs of the theorem above and references to other works that have proofs, see Melman 2021 [1].