My problem: is there a theorem which can easily check the existence of right half plane zeros in a polynomial with real coefficients: $p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0$, where ${a_n,a_{n-1},...,a_0} $ are all real coefficients.

I know under this setup the polynomial will have positive/negative real zeros and conjugate complex zeros. I wonder if there is a property that the polynomial can satisfy to avoid the existence of right half plane zeros.

Many thanks!

  • $\begingroup$ If $x=u+v\,i$ is a zero then $x^2-2ux+u^2+v^2$ is a divisor so a condition on the $a_k$ is needed which guarantees that if $x^2+sx+t^2$ is a divisor, then $s\ge0$. $\endgroup$ – John Wayland Bales Sep 12 '16 at 19:18
  • $\begingroup$ But we don't know what $s$ is if only $a_i$ given, correct? $\endgroup$ – Albert Sep 12 '16 at 19:42
  • $\begingroup$ One must know all the coefficients of $p(x)$. Then the requirement that a zero $x=u+v\,i$ of $p(x)$ satisfy $u\le0$ requires that any quadratic divisor $x^2+sx+t^2$ of $p(x)$ satisfy $s\ge0$. For $p(x)$ quadratic this requires only that all the coefficients have the same sign. For $p(x)$ of third degree the situation becomes more complicated since the value of $t^2$ cannot be neglected. I do not know the answer to your question but since no one was responding I decided to try to make a perhaps useful observation. $\endgroup$ – John Wayland Bales Sep 12 '16 at 19:54
  • $\begingroup$ I see. Anyway thanks! $\endgroup$ – Albert Sep 12 '16 at 20:33
  • $\begingroup$ Lookup stable polynomial and the Routh–Hurwitz theorem. $\endgroup$ – dxiv Sep 13 '16 at 1:33

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