Let $p_n$ be a polynomial of exactly degree $n$, with positive leading coefficient, and suppose that it has $n$ simple real roots. Let $y_1<\dots <y_{n+1}$ be real (simple) roots of $p_{n+1}$. Assume that $p_n/p_{n+1}$ is decreasing in each interval free of zeros of $p_{n+1}$. Then we must have $\lim_{x\to y_i^{\pm}}p_n(x)/p_{n+1}(x)=\pm \infty$. Why can we conclude from this that $p_n$ has exactly one root between two consecutive roots of $p_{n+1}$?
I was thinking about using the Intermediate Value Theorem, but the interval, for example, $(y_1,y_2)$ is not closed. What to do then?
Best Answer
Fix $n$ and put $f=p_n/p_{n+1}$. You have observed that $\lim_{x\to y_i^{\pm}}f(x)=\pm \infty $. Hence given $i$, there exists $a>y_i$ and $b<y_{i+1}$ such that $f(a)>0$ and $f(b)<0$. Furthermore, $a$ and $b$ can be chosen such that \begin{align} a-y_i&<\frac{y_{i+1}-y_i}{2}\, \\ y_{i+1}-b&<\frac{y_{i+1}-y_i}{2}\,. \end{align} Then $y_i<a<b<y_{i+1}$. By the intermediate value theorem, $f(r)=0$ for some $r\in(a,b)$.
Suppose, for contradiction, that $p_n$ has two distinct roots $r_1<r_2$ between $y_i$ and $y_{i+1}$. Then $f(r_1)=f(r_2)=0$. Since $f$ is decreasing, it must be that $p_n(x)=0$ for all $x\in[r_1,r_2]$, which is impossible.