1 Introduction

For a positive integer number n we consider

$$\begin{aligned} f(x)=a_n x^n + a_{n-1} x^{n-1} + \cdots +a_1 x +a_0 \ \ \ \ \mathrm{with} \ \ \ a_0,\ldots ,a_n>0. \end{aligned}$$
(1)

Let \(\mathbb {R}^+[n]\) be the family of all polynomials of the form (1). The polynomial

$$\begin{aligned} f^{[p]}(x):=a_n^p x^n + a_{n-1}^p x^{n-1} + \cdots +a_1^p x +a_0^p \end{aligned}$$
(2)

where \(p\in \mathbb {R}\) is called the pth Hadamard power of \(f\in \mathbb {R}^+[n]\). We say that the polynomial f with real coefficients is stable (f is a Hurwitz polynomial) if every zero of f has strictly negative real part. A necessary condition for a polynomial f with real coefficients to be stable is that f has all coefficients of the same sign. Let \(H_n\) be the family of all stable polynomials of degree n with positive coefficients.

In 1996 J.Garloff and D.G.Wagner proved in [1] that \(f\,\in \, H_n\) implies \(f^{[p]}\,\in \, H_n\) for all \(p\,\in \,\{1,2,3,\ldots \}\). The natural question arises of a set of real numbers p for which \(f^{[p]}\) is stable where \(f\in \mathbb {R}^+[n]\). We give some conditions on p and on f for \(f^{[p]}\) to belong to \(H_n\). Moreover, we show that \(f^{[p]}\) does not need to be stable for a stable polynomial f and an exponent \(p> 1\), contrary to Theorem 5 in [2].

Observe that if \(n=1\) or \(n=2\) then \(f^{[p]}\) is stable for every \(p\in \mathbb {R}\) and for all polynomials \(f\in \mathbb {R}^+[n]\). The case of \(n\ge 3\) is much more complicated, e.g., for \(f(x)=x^3+x^2+x+1\) we have \(f^{[p]}\not \in H_n\) for any \(p\in \mathbb {R}\). Therefore, we will consider only the case \(n\ge 3\).

1.1 Basic information

For relevant background material concerning Hurwitz polynomials and related topics see [5, Sec.11]. We list below selected theorems that will be useful in the paper.

The Hurwitz matrix H(f) associated to the polynomial \(f\in \mathbb {R}^+[n]\) is given as follows

$$\begin{aligned} H(f):= \left[ \begin{array}{ccccccc} a_{n-1} &{}\quad a_{n-3} &{}\quad a_{n-5} &{}\quad a_{n-7} &{}\quad \ldots &{} 0 \\ a_{n} &{}\quad a_{n-2} &{}\quad a_{n-4} &{}\quad a_{n-6} &{}\quad \ldots &{} 0 \\ 0 &{}\quad a_{n-1} &{}\quad a_{n-3} &{}\quad a_{n-5} &{}\quad \ldots &{} 0 \\ 0 &{}\quad a_{n} &{}\quad a_{n-2} &{}\quad a_{n-4} &{}\quad \ldots &{} 0 \\ 0 &{}\quad 0 &{}\quad a_{n-1} &{}\quad a_{n-3} &{}\quad \ldots &{} 0 \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{} \vdots \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots &{}\quad a_0 \end{array} \right] \in \mathbb {R}^{n\times n}. \end{aligned}$$

Denote by \(D_i(p)\) for \(i=1,\ldots ,n\) the ith leading principal minor of the Hurwitz matrix \(H(f^{[p]})\), i.e.,

$$\begin{aligned} D_1(p)=a_{n-1}^p, \ \ \ D_2(p)= \mathrm{det}\, \left[ \begin{array}{cc} a_{n-1}^p &{}\quad a_{n-3}^p \\ a_{n}^p &{}\quad a_{n-2}^p \\ \end{array} \right] , \ \ldots \ , \ D_n(p)=\det H(f^{[p]}). \end{aligned}$$

To simplify the writing, we put \(D_i:=D_i(1)\).

Theorem 1

Routh–Hurwitz criterion

If \(f\in \mathbb {R}^+[n]\) then \(f\in H_n\) if and only if \(D_i>0\) for all \(i=1,\ldots ,n\).

Theorem 2

(see [3, Th.2 and (1.10)])

If \(f\in H_n\) with \(n\ge 3\) then

$$\begin{aligned} \mathrm{det}\, \left[ \begin{array}{cc} a_{n-i} &{}\quad a_{n-i-2} \\ a_{n-i+1} &{}\quad a_{n-i-1} \\ \end{array} \right] >0 \ \ \ \ \ { for \ \ all} \ \ \ i=1,\ldots ,n-2. \end{aligned}$$

Theorem 3

(see [4])

Let \(f\in \mathbb {R}^+[n]\) with \(n\ge 5\) and \(\gamma \) be the unique real root of the equation

$$\begin{aligned} \gamma (\gamma +1)^2=1. \end{aligned}$$

If \(\gamma \, a_{n-i}\,a_{n-i-1}>a_{n-i+1}\,a_{n-i-2}\) for every \(i=1,\ldots ,n-2\) then \(f\in H_n\).

1.2 Counterexample

Theorem 5 in [2] asserts that if \(f\in H_n\) then \(f^{[p]}\in H_n\) for all \(p\ge 1\). We construct below a counterexample to this statement.

For a fixed polynomial \(f\in \mathbb {R}^+[n]\) with \(n\ge 3\) consider the following decomposition

$$\begin{aligned} f(x)=g(x^2)+x\,h(x^2), \text {where} \,g\, \text {and} \,h\, \text {are polynomials of positive coefficients} \end{aligned}$$
(3)

It may be worth reminding the reader that g and h are called interlacing if

  • all zeros of g and h are real, negative and distinct,

  • between every two zeros of g there exists a zero of h and vice versa.

Among variants of Hermite–Biehler theorem we will apply the following one to construct a counterexample.

Theorem 4

(see [5, Chapter 6.3]) Every polynomial \(f\in \mathbb {R}^+[n]\) decomposed as in (3) is stable if and only if g and h are interlacing.

Counterexample 1

Let

$$\begin{aligned} g(t)=t^4+46\,t^3+791\,t^2+6026\,t+17160=(t+10)(t+11)(t+12)(t+13). \end{aligned}$$

Y. Wang and B. Zhang considered g in [6] and observed that for \(p=1.139\) the polynomial \(g^{[p]}\) has two nonreal zeros: \(-16.0617\pm 0.178468\,i\) (approximated value). Take now

$$\begin{aligned} h(t)=t^3+34.5\,t^2+395.75\,t+1509.375=(t+10.5)(t+11.5)(t+12.5) \end{aligned}$$

and put

$$\begin{aligned} f(x)= & {} g(x^2)+xh(x^2) \\= & {} x^8+x^7+46\,x^6+34.5\,x^5+791\,x^4+395.75\,x^3 \\ \quad&+6026\,x^2+1509.375\,x+17160. \end{aligned}$$

It is easy to verify that f is stable (e.g., by the Routh–Hurwitz criterion). We have \(f^{[p]}(x)=g^{[p]}(x^2)+x\,h^{[p]}(x^2)\) and thus, by Theorem 4 the polynomial \(f^{[p]}\) is not stable for \(p=1.139\). By means of Wolfram Mathematica 10.4 we found two zeros of \(f^{[1.139]}\) that have positive real part: \(0.00179025\pm 4.01279\,i\) (approximated value).

2 Main results

Now we will state and prove some sufficient conditions for \(f^{[p]}\) to be a Hurwitz polynomial for all \(p> p_0\) or \(p< p_1\) with some positive \(p_0\) and negative \(p_1\) depending only on coefficients of f. The polynomial f is assumed to be of the form (1) but need not to be stable. We will discuss separately three cases: \(n=3\), \(n=4\) and \(n\ge 5\). We start with a lemma and some necessary conditions for the Hurwitz stability.

2.1 Notations and preliminary results

For a fixed polynomial \(f\in \mathbb {R}^+[n]\) with \(n\ge 3\) and \(p\in \mathbb {R}\) we put

$$\begin{aligned} w_i(p):= a_{n-i-1}^p \, a_{n-i}^p - a_{n-i-2}^p \, a_{n-i+1}^p, \ \ \ \ \ i=1,\ldots ,n-2. \end{aligned}$$
(4)

Moreover, for ease of notation, throughout the paper we write \(w_i\) for \(w_i(1)\). Let

$$\begin{aligned} \overline{d}:= & {} \max _{1\le i\le n-2} \frac{a_{n-i-2} a_{n-i+1}}{a_{n-i-1} a_{n-i}},\\ \underline{d}:= & {} \min _{1\le i\le n-2} \frac{a_{n-i-2} a_{n-i+1}}{a_{n-i-1} a_{n-i}}. \end{aligned}$$

It is worth noticing that

  • if \(w_i>0\) for all i then \(\overline{d}<1\),

  • if \(w_i<0\) for all i then \(\underline{d}>1\).

Lemma 1

Let \(\lambda \in (0,1)\) and \(f\in \mathbb {R}^+[n]\) with \(n\ge 3\). Put

$$\begin{aligned} p_0:=\frac{\log \lambda }{\log \overline{d}}, \ \ \ \ \ p_1:=\frac{\log \lambda }{\log \underline{d}}. \end{aligned}$$
  1. 1.

    If \(w_i>0\) for all \(i=1,\ldots ,n-2\)

    $$\begin{aligned}&then \ \ \ \ \lambda a_{n-i-1}^p \, a_{n-i}^p - a_{n-i-2}^p \, a_{n-i+1}^p> 0 \ \ \ \ {for \ all} \\ \qquad&i=1,\ldots ,n-2 \ \ \ and \ \ \ p>p_0>0.\ \ \end{aligned}$$
  2. 2.

    If \(w_i<0\) for all \(i=1,\ldots ,n-2\)

    $$\begin{aligned}&then \ \ \ \ \lambda a_{n-i-1}^p \, a_{n-i}^p - a_{n-i-2}^p \, a_{n-i+1}^p > 0 \ \ \ \ {for \ all} \\ \qquad&i=1,\ldots ,n-2 \ \ \ and \ \ \ p<p_1<0.\ \ \end{aligned}$$

Proof

Firstly we show statement 1. Since \(w_i>0\) for all i, it follows that \(\overline{d}<1\) and hence for a fixed \(p>p_0\) we have \(\overline{d}^p<\lambda \). From the definition of \(\overline{d}\) we can easily conclude that \(\overline{d}a_{n-i-1} a_{n-i} \ge a_{n-i-2} a_{n-i+1}\) for all i and so

$$\begin{aligned} 0 \ \le \ \overline{d}^p a_{n-i-1}^p a_{n-i}^p - a_{n-i-2}^p a_{n-i+1}^p \ < \ \lambda a_{n-i-1}^p a_{n-i}^p - a_{n-i-2}^p a_{n-i+1}^p. \end{aligned}$$

In an analogous manner we can prove statement 2. Indeed, in this case we have \(\underline{d}>1\) and \(\underline{d}^p<\lambda \) for \(p<p_1\). From the definition of \(\underline{d}\) we get \(\underline{d}\, a_{n-i-1} \, a_{n-i} \le a_{n-i-2} \, a_{n-i+1}\) for all i. Hence

$$\begin{aligned} 0 \ \le \ \underline{d}^p \, a_{n-i-1}^p \, a_{n-i}^p - a_{n-i-2}^p \, a_{n-i+1}^p \ < \ \lambda \, a_{n-i-1}^p \, a_{n-i}^p - a_{n-i-2}^p \, a_{n-i+1}^p \end{aligned}$$

and the proof is completed.\(\square \)

We give below some sufficient conditions for \(f^{[p]}\) not to be stable. This is a direct consequence of Theorem 2.

Theorem 5

Let \(f\in \mathbb {R}^+[n]\) with \(n\ge 3\).

  1. 1.

    If \(w_i\ge 0\) for some \(i\in \{1,\ldots ,n-2\}\) then \(f^{[p]}\not \in H_n\) for all \(p\le 0\).

  2. 2.

    If \(w_i\le 0\) for some \(i\in \{1,\ldots ,n-2\}\) then \(f^{[p]}\not \in H_n\) for all \(p\ge 0\).

2.2 Case \(n=3\)

In this subsection we consider \(f(x)=a_3\ x^3 + a_2\ x^2 + a_1\ x + a_0\) with positive coefficients \(a_3,\, a_2,\, a_1, \, a_0\). For \(n=3\) the family of \(w_i\)’s [see (4)] is reduced to the unique element \(w_1=a_1\, a_2-a_0\, a_3\).

Theorem 6

For any polynomial \(f\in \mathbb {R}^+[n]\) with \(n=3\) we have

  1. 1.

    If \(w_1>0\) then \(f^{[p]}\in H_3\) for all \(p> 0\).

  2. 2.

    If \(w_1<0\) then \(f^{[p]}\in H_3\) for all \(p< 0\).

Proof

In order to prove statement 1, we observe that \(w_1>0\) implies \(w_1(p)=a_1^p\, a_2^p-a_0^p\, a_3^p>0\) for every \(p>0\). By the Routh–Hurwitz criterion we get the stability of \(f^{[p]}\) for \(p>0\), because

$$\begin{aligned} D_1(p)=a_2^p>0, \ \ \ D_2(p)=w_1(p) \ \ \ \hbox {and} \ \ \ D_3(p)=a_0^pw_1(p). \end{aligned}$$

In an analogous manner we can prove statement 2. \(\square \)

2.3 Case \(n=4\)

We start this subsection with a simple characterization of stable polynomials of degree 4 with positive coefficients.

Proposition 7

Let \(f\in \mathbb {R}^+[n]\) with \(n=4\). The polynomial f is stable if and only if

$$\begin{aligned} \frac{a_1\, a_4}{a_2\, a_3} + \frac{a_0\, a_3}{a_1\, a_2} \ < \ 1. \end{aligned}$$
(5)

Proof

It is easily computed that

$$\begin{aligned} D_1=a_3, \ \ \ \ D_2=a_2\, a_3-a_1\, a_4, \ \ \ \ D_3=a_1\, a_2\, a_3-a_0\, a_3^2-a_1^2\, a_4, \ \ \ \ D_4=a_0\, D_3. \end{aligned}$$

By the Routh–Hurwitz criterion, \(f\in H_4\) implies \(D_3>0\), i.e.,

$$\begin{aligned} a_1 a_2 a_3 \ > \ a_0 a_3^2+a_1^2 a_4. \end{aligned}$$

Dividing by \(a_1\, a_2\, a_3\) we obtain inequality (5).

For the reverse implication, we can conclude from (5) that

$$\begin{aligned} \frac{a_1\, a_4}{a_2\, a_3} \ < \ 1 \end{aligned}$$

and hence \(D_2>0\). Moreover, an immediate consequence of (5) is \(D_3>0\), and so \(D_4>0\). Once again we use the Routh–Hurwitz criterion and get the stability of f. \(\square \)

Note that for \(n=4\) and any function \(f(x)=a_4 \ x^4 + a_3\ x^3 + a_2\ x^2 + a_1\ x + a_0\) we have only two \(w_i\)’s defined by (4):

$$\begin{aligned} w_1=a_2 a_3-a_1 a_4, \ \ \ \ \ \ w_2=a_1 a_2-a_0 a_3 \end{aligned}$$

and

$$\begin{aligned} \overline{d}:=\max \left\{ \frac{a_1 a_4}{a_2 a_3}, \ \frac{a_0 a_3}{a_1 a_2}\right\} \ \ \ \ \ \ \underline{d}:=\min \left\{ \frac{a_1 a_4}{a_2 a_3}, \ \frac{a_0 a_3}{a_1 a_2}\right\} . \end{aligned}$$

It is worth recalling from the beginning of Sect. 2.1 that for f with positive coefficients we have \(\overline{d}< 1\) if all \(w_i\)’s are positive and \(\underline{d}> 1\) whenever all \(w_i\)’s are negative.

Theorem 8

Let \(f\in \mathbb {R}^+[n]\) with \(n=4\) and

$$\begin{aligned} p_0:=\frac{\log 0.5}{\log \overline{d}} \ \ \ \ \ \ p_1:=\frac{\log 0.5}{\log \underline{d}}. \end{aligned}$$
  1. 1.

    If \(w_1, \, w_2>0\) then \(f^{[p]}\in H_4\) for all \(p> p_0> 0\).

  2. 2.

    If \(w_1, \, w_2<0\) then \(f^{[p]}\in H_4\) for all \(p< p_1<0\).

Moreover, the constants \(p_0\) and \(p_1\) are the best possible, i.e., for \(p_0\) it means that there exists a polynomial f of degree 4 with positive coefficients and \(w_1,w_2>0\) such that \(f^{[p]}\) is not stable for every \(p\le p_0\).

Proof

For the proof of statement 1, we use Lemma 1. For \(\lambda =1/2\) and \(p>p_0\) we have

$$\begin{aligned} \frac{1}{2}\, a_2^p\, a_3^p-a_1^p\, a_4^p\> \ 0, \ \ \ \ \ \ \frac{1}{2} \, a_1^p\, a_2^p - a_0^p\, a_3^p \ > \ 0. \end{aligned}$$

Consequently,

$$\begin{aligned} \frac{a_1^p\, a_4^p}{a_2^p\, a_3^p}<\frac{1}{2} \ \ \ \ \mathrm{and} \ \ \ \ \frac{a_0^p\, a_3^p}{a_1^p\, a_2^p} < \frac{1}{2} \end{aligned}$$

and therefore,

$$\begin{aligned} \frac{a_1^p\, a_4^p}{a_2^p\, a_3^p} + \frac{a_0^p\, a_3^p}{a_1^p\, a_2^p} \ < \ 1. \end{aligned}$$
(6)

By Proposition 7 we get the stability of \(f^{[p]}\) for \(p>p_0\). Statement 2 can be proved in an analogous fashion.

By Example 2 given below we show that the constants \(p_0\) and \(p_1\) cannot be improved. \(\square \)

Example 2

Consider the polynomial

$$\begin{aligned} f(x)=2x^4+x^3+5x^2+x+2. \end{aligned}$$

In this case we have

$$\begin{aligned} w_1=5\cdot 1-1\cdot 2=3>0, \ \ \ \ \ \ \ w_2=1\cdot 5-2\cdot 1=3>0 \end{aligned}$$

and

$$\begin{aligned} \overline{d}=\max \left\{ \frac{2}{5},\frac{2}{5}\right\} =0.4, \ \ \ \ \ \ \ \ \ \ \ p_0=\frac{\log 0.5}{\log 0.4}. \end{aligned}$$

Fix \(p\le p_0\). By Proposition 7, \(f^{[p]}\in H_4\) if and only if inequality (6) holds. We calculate

$$\begin{aligned} \frac{a_1^p\, a_4^p}{a_2^p\, a_3^p} + \frac{a_0^p\, a_3^p}{a_1^p\, a_2^p}= & {} \left( \frac{a_1\, a_4}{a_2\, a_3}\right) ^p + \left( \frac{a_0\, a_3}{a_1\, a_2}\right) ^p=(0.4)^p+(0.4)^p \\= & {} 2\cdot (0.4)^p\ge 2\cdot (0.4)^{p_0}=2\cdot 0.5=1. \end{aligned}$$

We see that inequality (6) does not hold and consequently \(f^{[p]}\) is not stable. Additionally, we can easily verify by Proposition 7 that polynomial f is stable.

Corollary 9

If \(f\in H_4\) then \(f^{[p]}\in H_4\) for all \(p\ge 1\).

Proof

Since \((t^p+s^p)^{1/p} \le t+s\) for all \(s,t\ge 0\) and \(p\ge 1\), we have

$$\begin{aligned} \frac{a_1^p\, a_4^p}{a_2^p\, a_3^p} + \frac{a_0^p\, a_3^p}{a_1^p\, a_2^p} \ \le \ \left( \frac{a_1\, a_4}{a_2\, a_3} + \frac{a_0\, a_3}{a_1\, a_2}\right) ^p \ < \ 1 \end{aligned}$$

the last estimate being a consequence of the stability of f and Proposition 7. Once again we use Proposition 7 and we get the stability of \(f^{[p]}\). \(\square \)

2.4 Case \(n\ge 5\)

The main result of this subsection will be based on Theorem 3 that deals with \(n\ge 5\). We remind the reader that \(\gamma \) denotes the unique real root of the equation \(\gamma (\gamma +1)^2=1\). One can verify that \(\gamma \in (0.4655,0.466)\). Quantities \(w_1,\ldots ,w_{n-2}\) and \(\overline{d}, \underline{d}\) have been defined in the beginning of Sect. 2.1.

Theorem 10

Let \(f\in \mathbb {R}^+[n]\) with \(n\ge 5\) and

$$\begin{aligned} p_0:=\frac{\log \gamma }{\log \overline{d}} \ \ \ \ \ \ p_1:=\frac{\log \gamma }{\log \underline{d}}. \end{aligned}$$
  1. 1.

    If \(w_1,\ldots , w_{n-2}>0\) then \(f^{[p]}\in H_n\) for all \(p> p_0> 0\).

  2. 2.

    If \(w_1,\ldots ,w_{n-2}<0\) then \(f^{[p]}\in H_n\) for all \(p< p_1<0\).

Proof

Take \(p>p_0\) in the case of \(w_1,\ldots , w_{n-2}>0\) or \(p<p_1\) in the case \(w_1,\ldots ,w_{n-2}<0\). In both cases, by Lemma 1 used for \(\lambda =\gamma \), we have \(\gamma a_{n-i-1}^p \, a_{n-i}^p - a_{n-i-2}^p \, a_{n-i+1}^p > 0\) for all \(i=1,\ldots ,n-2\). Thanks to Theorem 3 we obtain the stability of \(f^{[p]}\) and the proof is completed. \(\square \)

Let us observe that \(p_0\) and \(p_1\) in Theorem 10 are not far from being optimal as evidenced in the next example.

Example 3

Consider the polynomial

$$\begin{aligned} f(x)=x^5+5x^4+2x^3+5x^2+x+1. \end{aligned}$$

We have

$$\begin{aligned} w_1= & {} a_3\, a_4-a_2\,a_5=5>0, \ \ \ \ w_2=a_2\, a_3-a_1\,a_4=5>0, \ \ \ \ \\ w_3= & {} a_1\,a_2-a_0\,a_3=3>0 \end{aligned}$$

and

$$\begin{aligned} \overline{d}=\max \left\{ \frac{a_2\,a_5}{a_3\, a_4}, \frac{a_1\,a_4}{a_2\, a_3}, \frac{a_0\,a_3}{a_1\,a_2}\right\} = \max \left\{ \frac{1}{2},\frac{1}{2},\frac{2}{5}\right\} =\frac{1}{2}. \end{aligned}$$

The Hurwitz matrix H(f) associated to f is

$$\begin{aligned} H(f)= \left[ \begin{array}{ccccc} 5 &{}\quad 5 &{}\quad 1 &{}\quad 0 &{}\quad 0 \\ 1 &{}\quad 2 &{}\quad 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 5 &{}\quad 5 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 2 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 5 &{}\quad 5 &{}\quad 1 \end{array} \right] . \end{aligned}$$

The leading principal minors are

$$\begin{aligned} D_1=5, \ \ \ \ D_2=5, \ \ \ \ D_3=5, \ \ \ \ D_4=-1<0, \ \ \ \ D_5=-1<0 \end{aligned}$$

and therefore, by the Routh–Hurwitz criterion, f is not stable.

Now take \(p\in \mathbb {R}\) and compute the 4th leading principal minors of \(H(f^{[p]})\):

$$\begin{aligned} D_4(p)=50^p+5^p-25^p-25^p-20^p-1+5^p+10^p. \end{aligned}$$

If we take p close to 1 then \(f^{[p]}\) is not stable because of the continuity of exponential functions and since \(D_4(1)<0\).

On the other hand, by Theorem 10, \(f^{[p]}\) is stable for all \(p\ge 1.1032\) as

$$\begin{aligned} p_0=\frac{\log \gamma }{\log \overline{d}}=\frac{\log \gamma }{\log 0.5}< \frac{-\log 0.4655}{\log 2}\approx 1.10315 < 1.1032. \end{aligned}$$

We conclude that the quantity \(p_0\) given in Theorem 10 is close to the value, where the stability of \(f^{[p]}\) changes.

The above example shows also that Theorem 8 proved for \(n=4\) cannot be applied for polynomials of degree 5, because by Theorem 8 we get \(f^{[p]}\in H_n\) for all \(p>\frac{\log 0.5}{\log \overline{d}}\). However, for the polynomial f considered in Example 3 we have \(\frac{\log 0.5}{\log \overline{d}}=1\) and we see that \(f^{[p]}\) is not stable for p close to 1.

We can show by the next example that the constant \(\gamma \) in Theorem 3 is close to the optimal one.

Example 4

Let

$$\begin{aligned} f(x)=x^5+5x^4+\left( 3-\tfrac{2}{\sqrt{5}}\right) \, x^3+5x^2+x+1. \end{aligned}$$

Observe that f has all positive coefficients and for

$$\begin{aligned} \lambda \ = \ 0.475 \ > \ \frac{1}{3-\frac{2}{\sqrt{5}}} \approx 0.47493 \end{aligned}$$

that is close to \(\gamma \in (0.4655,0.466)\), we have

$$\begin{aligned}&\lambda a_3\, a_4-a_2\,a_5=\lambda \left( 3-\tfrac{2}{\sqrt{5}}\right) \cdot 5 -5>0,\\&\quad \lambda a_2\, a_3-a_1\,a_4=\lambda \cdot 5\left( 3-\tfrac{2}{\sqrt{5}}\right) -5>0,\\&\quad \lambda a_1\,a_2-a_0\,a_3=\lambda \cdot 5 - \left( 3-\tfrac{2}{\sqrt{5}}\right)> \frac{5}{\left( 3-\tfrac{2}{\sqrt{5}}\right) }\\&\quad -\left( 3-\tfrac{2}{\sqrt{5}}\right) =\frac{12}{5\left( 3-\tfrac{2}{\sqrt{5}}\right) }(\sqrt{5}-2) > 0. \end{aligned}$$

By Theorem 3 analogous inequalities satisfied for \(\gamma \) (instead of \(\lambda \)) imply the Hurwitz stability of f. However, in the considered case we get

$$\begin{aligned} D_4= \det \left[ \begin{array}{cccc} 5 &{}\quad 5 &{}\quad 1 &{}\quad 0 \\ 1 &{}\quad 3-\frac{2}{\sqrt{5}} &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 5 &{}\quad 5 &{}\quad 1 \\ 0 &{}\quad 1 &{}\quad 3-\frac{2}{\sqrt{5}} &{}\quad 1 \\ \end{array} \right] =0 \end{aligned}$$

and therefore, by the Routh–Hurwitz criterion f is not stable.