Newsgroups: sci.math.research
From: sawyer@ramsey.cs.laurentian.ca (P. Sawyer)
Subject: elementary symmetric polynomials
Date: Tue, 11 Apr 1995 18:13:33 GMT

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Consider $x_1$, \dots, $x_n$ such that for all $i$, $j$,  
$|\Im(x_i-x_j)|<a$  ($\Im$ is the imaginary part) where  
$a>0$.  Suppose $p_1$, \dots, $p_{n-1}$ are the elementary 
symmetric polynomials in $n-1$ variables ($p_k(y_1,\dots,y_{n-1})
=\sum_{i_1<\dots<i_k}\,y_{i_1}\cdots y_{i_k}$).  
Let $t_1\geq 0$, \dots, $t_n\geq0 $ be such that 
$t_1+\dots+t_n=1$.

Define $y_1$, \dots, $y_{n-1}$ (up to order) by
$p_k(y_1,\dots,y_k)=\sum_{j=1}^{n}\,t_j\,p(x^j)$ where $x^j=(x_1,\dots,x_{j-1},x_{j+1},\dots, x_n)$.  

The question: can we say that $|\Im(y_i-y_j)|<a$ for 
all $i$, $j$~?

\end{document}

Patrice Sawyer
Dept. of Math. & Comp. Sci.
Laurentian University
sawyer@ramsey.cs.laurentian.ca

==============================================================================

Date: Wed, 12 Apr 95 14:52:22 CDT
From: rusin (Dave Rusin)
To: sawyer@ramsey.cs.laurentian.ca
Subject: Re: elementary symmetric polynomials
Newsgroups: sci.math.research

In article <3megsd$aos@ramsey.cs.laurentian.ca> you write:
>Define $y_1$, \dots, $y_{n-1}$ (up to order) by
>$p_k(y_1,\dots,y_k)=\sum_{j=1}^{n}\,t_j\,p(x^j)$ where 
                ^^^                       ^^^^^^
I'm guessing you mean   y_{n-1}  and   p_k(x^j)   here.
In that case, I get that the  y_i  are the zeros of the rational function
F(Y)=Sum  t_j/(Y-x_j).  Is that what you mean?

I would assume that the way to prove this is to show that the  y_i  all
like in the convex hull of the  x_j. Do you already know if that stronger
statement is false?

dave

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Date: Wed, 12 Apr 95 15:56:25 -0400
From: Patrice Sawyer <sawyer@lie.cs.laurentian.ca>
To: rusin@math.niu.edu (Dave Rusin)
Subject: Re: elementary symmetric polynomials

Hi Dave,

I think I have the result you are suggesting:

\begin{lemma}
Suppose $z_1$, \dots, $z_n\in \bold{C}$.  Fix $t_1\geq0$, \dots,  
$t_n\geq 0$ such that $\sum_{i=1}^{n}\,t_i=1$.  Let
$$
P(z)=\sum_{i=1}^{n}\,t_i\,\prod_{j\not=i}\,(z-z_j).
$$
Then the zeros of $P$ belong to the convex hull of $z_1$, \dots,  
$z_n$.
\end{lemma}

\begin{pf}
Let $Q(z)=\prod_{i=1}^{n}\,(z-z_j)$.  We have
$$
\frac{P(z)}{Q(z)}
=\sum_{i=1}^{n}\,\frac{t_i}{z-z_i}
= \sum_{i=1}^{n}\,\frac{t_i}{|z-z_i|^2}\,(\bar{z}-\bar{z}_i).
$$
If $z$ is a zero of $P$ but not a zero of $Q$ then
$$
0=\sum_{i=1}^{n}\,\frac{t_i}{|z-z_i|^2}\,(\bar{z}-\bar{z}_i)
$$
and, taking  conjugates,
$$
 \sum_{i=1}^{n}\,\frac{t_i}{|z-z_i|^2}\,z
=\sum_{i=1}^{n}\,\frac{t_i}{|z-z_i|^2}\,z_i
$$
which implies the result.
\end{pf}


What do you think ?

Patrice
[sig deleted]

==============================================================================
Addendum - 10/14/1997 -- djr

John Brillhart recalls a result of Gauss: if  F  is a polynomial with
complex coefficients, the roots of  F'  lie in the convex hull of the
roots of  F. Hence if  F  is the reciprocal of a polynomial, the
roots of  F'  lie in the convex hull of the poles of  F. What about
if  F  is a general rational function; do the roots of  F'  lie in the
convex hull of the roots and poles of  F? (Clearly the poles of F'
are the same as the pole of  F).

The answer is no; a counterexample is  F(z) = z/(z^3-1).

What can be said of the location of the roots of  F'  when  F   is rational?