1
$\begingroup$

Consider the following equation (in variable $x$) for $\lambda_i\geq 0$

$$ p(x)=\frac{w_1}{(x-\lambda_1)}+\frac{w_2}{(x-\lambda_2)}\cdots+\frac{w_n}{(x-\lambda_n)} $$

I am interested in characterizing the roots of the above equation. Is there any body of literature which I can refer to?

Specifically, I am interested bounding the sum of top $k$ roots of the above polynomial for the case when $\sum_{i=1}^nw_i^2 = 1$

$\endgroup$
6
  • 1
    $\begingroup$ I don't understand the question. $p(x)$ is not a polynomial... are you referring to its numerator? $\endgroup$ Sep 14, 2016 at 23:02
  • $\begingroup$ Yes. I am referring to the numerator of $p(x)$. I'll make an appropriate edit. I was abusing the notation a little for sake of clarity :). $\endgroup$ Sep 14, 2016 at 23:29
  • $\begingroup$ You might use the Argument Principle or Rouché's theorem to control the number of zeros of $p(x)$ in a region. $\endgroup$ Sep 14, 2016 at 23:31
  • 1
    $\begingroup$ Note that the location of the roots is unchanged for $cp(x)$ so the value of $\sum_1^nw_i^2$ is irrelevant. $\endgroup$ Sep 15, 2016 at 5:28
  • $\begingroup$ The coefficients $w_i$ are real numbers ? If yes, you know something on their signs ? $\endgroup$
    – user111
    Sep 16, 2016 at 8:29

2 Answers 2

5
$\begingroup$

We may assume that the $\lambda_i$ are distinct and that $\lambda_i \lt \lambda_j$ for $i \lt j$.

Given the $\lambda_i$ but no assumption on the $w_i,$ we can get the roots to be any set of $n-1$ points on the real line. In fact we can get it to be any set of $n-1$ or less distinct points except that the number is equal in parity to $n-1.$ So you need to be more specific. If the $w_i$ are all positive then the roots you seek interlace the $\lambda_i$ so the sum of the top $k$ roots is somewhere between $\sum_{n-k}^{n-1}\lambda_i$ and $\sum_{n-k+1}^{n}\lambda_i.$ These roots are also the places where $\prod|x-\lambda_i|^{m_i}$ has local maxima.

So perhaps for a reference you could look into interlacing of roots. However I don't have a good place for you to start.

The location of the roots is unchanged if we replace $p(x)$ with $cp(x)$ so we may assume $\sum w_i=1.$ The case $\sum w_i=0$ might be interesting , but I'll ignore it here.

The numerator of $$\sum_1^n\frac{w_i}{x-\lambda_i}=\frac{n(x)}{\prod(x-\lambda_i)}= \frac{x^{n-1}+\sum_0^{n-2}a_jx^j}{\prod(x-\lambda_i)} $$ is a monic polynomial of degree $n-1$.

Any choice of the $n-1$ values $w_1,w_2,\cdots,w_{n-1}$ (with $w_n=1-\sum_1^{n-1}w_i)$ uniquely determines the $n-1$ coefficients $a_0,a_1,\cdots,a_{n-2}.$ The converse is true as well, we can pick $n(x)$ to be any monic polynomial of degree $n-1$ and the $w_i$ are uniquely determined. It might be reasonable to specify that the $w_i$ are non-zero, but I won't in order to preserve this correspondence. From the way I specified the denominator, having a particular $w_i=0$ amounts to making $\lambda_i$ a root of $n(x).$

From now on I will assume that the $w_i$ are all positive and give two proofs that there is exactly one root in each interval $(\lambda_i,\lambda_{i+1})$

First proof: $p(x)$ is continuous in $(\lambda_{i},\lambda_{i+1}).$ Also, the one sided limits $$\lim_{x\to \lambda_i}p(x)=\pm\infty$$ with $+ \infty$ from above and $-\infty$ from below. This means the graph crosses the axis in each interval and there can only be $n-1$ roots.

Second proof: Consider $q(x)=\prod(|x-\lambda_i|)^{w_i}.$ It is continuous and non-negative touching the axis at each $\lambda_i$ but nowhere else. It has a local maximum in each $(\lambda_{i},\lambda_{i+1}).$ Also, $\ln |q(x)|$ has a local maximum at the same locations. Thus the derivative of $\ln|q(x)|$ has a zero in each interval (again at the same places). But that derivative is $p(x).$

$\endgroup$
4
  • $\begingroup$ Thanks! Any idea above mix of positive/negative $w_i$'s? $\endgroup$ Sep 16, 2016 at 20:30
  • 1
    $\begingroup$ I improved the answer and addressed that to some extent. $\endgroup$ Sep 17, 2016 at 4:29
  • $\begingroup$ @Aaron I am sorry, I do not understand why the numerator should not have repeated roots, what is wrong with $(1/6) / (x-1) - (1/2) / (x-3) + 4/3 / (x - 4) \ = \ (x-2)^2 /( (x-1)(x-3)(x-4) )$ ? $\endgroup$ Sep 18, 2016 at 9:57
  • $\begingroup$ You are right. I was worried about repeated roots in the denominator. Silly error. $\endgroup$ Sep 18, 2016 at 23:08
4
$\begingroup$

Concerning the literature, Morris Marden's "Geometry of Polynomials" from 1966 is a good place to start reading, for instance, the first chapter gives a number of different possible interpretations of the sums you consider.

Kenneth B. Stolarsky's review in

The American Mathematical Monthly Vol. 112, No. 7 (Aug. - Sep., 2005), pp. 664-671

of two more recent books on the subject contains a lot of additional information and is a fine read, reviewed books:

Analytic Theory of Polynomials by Qazi Ibadur Rahman, Gerhard Schmeisser;

Complex Polynomials by Terry Sheil-Small.

From the introduction of Marden's text, you can easily see, that the case

$$ \sum w_i \ = \ 0 \quad (*)$$

not explicitely adressed by Aaron Meyerowitz is an exceptional case you should take note of: Clearing denominators gives a polynomial on the right hand side, whose degree is lower than n-1 if and only if $(*)$ is true, so you will have less roots in that case. This is quite obvious in terms of physics, especially fluid dynamics:

The roots of $p(x)$ are the stagnation points of a flow which has sources and sinks, corresponding to positive and negative signs of the $w_i$ of strength $|w_i|$. If you have for example a unit source and a unit sink, there will be no stagnation points and hence no root, one less than expected, if $(*)$ does not vanish.

$\endgroup$
2
  • $\begingroup$ I had indeed been unaware, that the coefficients in the partial fraction decomposition of a polynomial $r(x)$ of degree at most $n-1$ with respect to a polynomial $s(x)$ of degree $n$ sum to zero if and only if $r(x)$ is of lower degree than $n-1$. And I just learned that the rule known in germany as Zuhalteregel is known as Heavyside cover-up in english speaking countries ... $\endgroup$ Sep 18, 2016 at 6:41
  • $\begingroup$ I have been sloppy in two places: As the polynomials $r(x)$ and $s(x)$ in $r(x)/s(x)$ are real valued, the number of roots maybe lower even in the case of the degree of $r(x)$ being $n-1$ and the roots of $s(x)$ should all be simple, if the vanishing of $(*)$ is to be equivalent with the degree of $r(x)$ being at most $n-2$. $\endgroup$ Sep 18, 2016 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.