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Suppose a fixed $n^{\text{th}}$ degree monic polynomial is given $$ p(x) = x^{n} + a_{n-1} x^{n-1} + \dots + a_0,$$ with coefficients vector $a = (a_{n-1}, \dots , a_0) \in \mathbb R^n$. Now we consider a parameterized family \begin{align*} p(x, r) = x^{n} + ra_{n-1} x^{n-1} + \dots + ra_0, \end{align*} with $r \in \mathbb R$.

Suppose for some $r_0 \in \mathbb R$, we have $p(i \alpha, r_0) = 0$, i.e., $p(x, r_0)$ has a zero on the imaginary axis. Let us further assume for some $\delta > 0$, we have $p(x, r)$ has all its zeros on the right half plane of $\mathbb C$ for each $r \in (r_0, r_0 + \delta)$.

My question is: is it possible that for all $r \in (r_0 - \delta, r_0)$ we also have $p(x, r)$ has all its zeros on the right half plane (we may include the imaginary axis). That is, is there a monic polynomial, that allows us to parametrize as above, such that the parametrized family has zeros touching the imaginary axis but all the zeros are confined in the right half plane when we vary $r$ continuously?

As commented by @saulspatz, $x^2 + r$ will have all its zeros on the imaginary axis if $r < 0$. I was in mind asking the case that at least there exists some $r \in (r_0 - \delta, r_0)$ such that some of the roots of $p(x, r)$ will move off the imaginary axis.

user1101010
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  • Doesn't $p(x)=x^2+1$ work, or am I missing something? – saulspatz Dec 05 '18 at 05:32
  • @saulspatz I don't think that works. What is the choice of $r_0$ such that the zeros move off the imaginary axis, to the right half-plane when you increase $r$ a bit? – Jyrki Lahtonen Dec 05 '18 at 05:39
  • @JyrkiLahtonen: I was reading your answer. Why did you delete it? I didn't see a problem of your answer. – user1101010 Dec 05 '18 at 05:41
  • @saulspatz: Thanks. Your comment fits in the setting. But I meant to make the initial choice of $a$ or at least existing some choice of $r$ such that $p(x, r)$ has all zeros on the open right half plane of $\mathbb C$. Will change the formulation of the problem. – user1101010 Dec 05 '18 at 05:45
  • If you allow the coefficients to be functions $a_i(r)$ other than linear, i.e. $a_i(r)=ra_i$, then my answer works. Gotta go and commute. Back in a few hours. – Jyrki Lahtonen Dec 05 '18 at 05:46
  • @JyrkiLahtonen: In the deleted answer of yours, with the consideration $a_i(r)$ not linear, you mentioned that the situation described in the problem would happen if the zeros on the imaginary axis are of multiplicity greater than $1$. Could you elaborate why it is the case? – user1101010 Dec 05 '18 at 07:21
  • @JyrkiLahtonen The roots stay on the imaginary axis, but this is specifically allowed in the question. The OP says, "we may include the imaginary axis." The OP has now clarified the question. – saulspatz Dec 05 '18 at 14:41

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I see it is essentially the same question as another one of yours: Can we have a simple root of real polynomials with coefficients as linear functions such that the curve of the root is tangent to imaginary axis?. For somebody who is curious about the result, I leave the link.

Myunghyun Song
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