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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.

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    $\begingroup$ I don't see why $P$ is a polynomial. $\endgroup$ – bubba Jan 20 '17 at 13:02
  • $\begingroup$ 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. $\endgroup$ – Elias Strehle Jan 20 '17 at 13:16
  • $\begingroup$ It's not true if $b_1,...,b_n$ are not $n$ distinct values. $\endgroup$ – DanielWainfleet Jan 20 '17 at 13:22
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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 $x$.

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  • $\begingroup$ This argument needs to be prefaced of course that WLOG we can assume: $b_1<b_2<b_3<...<b_n$ $\endgroup$ – dimpol Jan 20 '17 at 13:26
  • $\begingroup$ On second thought, I don't think this argument holds if the $b_i$ aren't all distinct $\endgroup$ – dimpol Jan 20 '17 at 13:27
  • $\begingroup$ That's true ... fortunately they are distinct in the problem I need to solve. Thanks for the great answer! $\endgroup$ – Elias Strehle Jan 20 '17 at 13:46
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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 $(b_i,b_{i+1})$.

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