2
$\begingroup$

I am trying to find some general properties of the zeros of

$P(z) = \sum_{i=1}^n \frac{\alpha_i}{z+z_i}$,

with $\sum_{i} \alpha_i = 0$, $z_i \in [-M\; 0], i=1,\ldots,n$ and all $\alpha_i$ and $z_i$ are real, and $M<\infty$.

Actually, what I really hope to find is that the real part of the zeros lie in the same interval as the poles, i.e., in the interval $[-M \; 0]$.

This problem arises in a particular type of dynamic systems over graphs. The function $P(z)$ is the transfer-function of a linear system running a consensus protocol, so the poles of the system are the eigenvalues of the combinatorial Laplacian.

Thanks!

Improve this question
$\endgroup$
7
  • $\begingroup$ The poles are $-z_i\in[0,M]$, not in $[-M,0]$. Or did you want to write $\sum_{i=1}^n\frac{\alpha_i}{z-z_i}$? $\endgroup$
    – Emil Jeřábek
    Commented Feb 25, 2011 at 11:23
  • $\begingroup$ Hi Emil, Yes, you are correct. $z_i \in [0, \; M]$ and the poles of the system are therefore in the interval $[-M, \; 0]$. $\endgroup$
    – dan
    Commented Feb 25, 2011 at 11:58
  • $\begingroup$ I received a couple comments (that appear deleted now) that suggest the Gauss-Lucas theorem might be applicable. I guess this is not the case, as the constants $\alpha_i$ are not integer (and may not even be rational). Are there any generalizations of the Gauss-Lucas theorem that can handle this case? $\endgroup$
    – dan
    Commented Feb 25, 2011 at 12:34
  • $\begingroup$ Sorry that was my comment earlier, and I deleted it because as you say it doesn't answer the question. I somehow read the condition as $\alpha_i\geq 0$. I think that by approximating $\alpha_i$ by rationals and rescaling makes it safe to assume that they are integers, bu this doesn't help as the only case I know how to put any sort of bound on the roots of $P$ is when $\sum \alpha_i \neq 0$. $\endgroup$
    – Gjergji Zaimi
    Commented Feb 25, 2011 at 13:09
  • $\begingroup$ Dan, are you still here? Does my answer of 27 February meet the specifications, or did I miss the point? $\endgroup$
    – Gerry Myerson
    Commented Mar 5, 2011 at 3:16

1 Answer 1

Reset to default
2
$\begingroup$

$${1\over z+1}+{-2\over z+10}+{1\over z+20}={170-z\over(z+1)(z+10)(z+20)}$$ seems to satisfy your conditions with $n=3$ and $M=20$ but the zero at 170 is not in the same interval as the poles.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .