# How to show that this polynomial has $n$ positive roots?

Let $n \in \mathbb{N}$ and $a_1, \dots, a_n, b_1, \dots, b_n > 0$. Define the function $$P(x) = \Bigg(\sum_{i=1}^n \frac{a_i}{x-b_i}\Bigg) - 1, \qquad x \in \mathbb{R}_+ \setminus \{b_1, \dots, b_n\}.$$

How can I check that $P$ has $n$ distinct positive roots? I have rearranged $P(x) = 0$ as $$\sum_{i=1} a_i \prod_{j \neq i} (x-b_j) = \prod_{i=1}^n (x-b_i),$$ but did not know how to proceed. I have also looked at Sturm chains without much success.

• I don't see why $P$ is a polynomial. – bubba Jan 20 '17 at 13:02
• You are right, of course. I changed that. The second equation, however, shows that the problem is equivalent to finding the roots of a polynomial of $n$-th degree. – Elias Strehle Jan 20 '17 at 13:16
• It's not true if $b_1,...,b_n$ are not $n$ distinct values. – DanielWainfleet Jan 20 '17 at 13:22

Take $x$ slightly larger than $b_i$, then the term $\frac{a_i}{x-b_i}$ is very large positive so $P$ is positive. Similarly take $x$ slightly less than $b_{i+1}$ and you get that $P$ is negetive. Since $P$ is continuous in each interval $(b_i,b_{i+1})$ there is a root in each such interval. That gives $n-1$ positive roots. There is one more root larger than $b_n$ since $P$ is negative for large $x$.
• This argument needs to be prefaced of course that WLOG we can assume: $b_1<b_2<b_3<...<b_n$ – dimpol Jan 20 '17 at 13:26
• On second thought, I don't think this argument holds if the $b_i$ aren't all distinct – dimpol Jan 20 '17 at 13:27
Let $b_i<b_{i+1}$ for $1\leq i\leq n-1.$ The derivative $P'(x)$ is negative wherever $P(x)$ is defined. So $P(x)$ is strictly decreasing on $(-\infty,b_1),$ and on $(b_i,b_{i+1})$ for $1\leq i \leq n-1,$ and on $(b_n, \infty).$
Observe that $\lim_{x\to \infty}P(x)=-1$ and $\lim_{x\to b_n^+}P(x)=+\infty,$ so $P(x)$ has a zero in $(b_n,\infty).$
Observe for $1<i\leq n-1$ that $\lim_{x\to b^-_{i+1}}P(x)=-\infty$ and that $\lim_{x\to b_i^+}P(x)=+\infty$ so $P(x)$ has a zero in $(b_i,b_{i+1})$.