Exactly one root of $p_n$ between two consecutive roots of $p_{n+1}$

Question

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?

Answer 1

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.

Answer 2

If we allow $p_n$ and $p_{n+1}$ to have common zeros, the claim is false.

As a counterexample, if we define $p_0,p_1,p_2,...$ recursively by $$ \left\lbrace \begin{align*} p_0\!&=1\\[4pt] p_{n+1}\!&=(x-n)p_n\\[4pt] \end{align*} \right. $$ then the hypothesis is satisfied but for all $n$, $p_n$ has no roots strictly between two consecutive roots of $p_{n+1}$.

So suppose we assume the additional condition that $p_n$ and $p_{n+1}$ have no common zeros.

Then the claim holds.

Proof:

Let $x_1 < \cdots < x_n$ be the roots of $p_n$.

Suppose $x_1 < y_1$.

On the interval $(-\infty,x_1)$, $p_n(x)/p_{n+1}(x)$ can't change sign, hence since one of $n,n+1$ is even and the other is odd, it follows that $p_n(x)/p_{n+1}(x)$ is negative on $(-\infty,x_1)$. But then since $x_1$ is a simple zero of $p_n$, the sign of $p_n(x)/p_{n+1}(x)$ must change from negative to positive as $x$ crosses $x_1$ from left to right, so $p_n/p_{n+1}$ is not decreasing the interval $(-\infty,y_1)$, contrary to hypothesis.

Hence $x_1 > y_1$.

Similarly, suppose $x_n > y_{n+1}$.

On the interval $(x_n,\infty)$, $p_n(x)/p_{n+1}(x)$ can't change sign, hence $p_n(x)/p_{n+1}(x)$ is positive on $(x_n,\infty)$. But then since $x_n$ is a simple zero of $p_n$, the sign of $p_n(x)/p_{n+1}(x)$ must change from negative to positive as $x$ crosses $x_n$ from left to right, so $p_n/p_{n+1}$ is not decreasing on the interval $(y_{n+1},\infty)$, contrary to hypothesis.

Hence $x_n < y_{n+1}$.

Next suppose that $p_n$ has more than one root strictly between some two consecutive roots of $p_{n+1}$.

Thus suppose $a < b < c < d$, where $a,d$ are consecutive roots of $p_{n+1}$ and $b,c$ are consecutive roots of $p_n$.

Since $p_n/p_{n+1}$ is decreasing on $(a,d)$, and $b,c$ are simple zeros of $p_n$, it follows that $p_n(x)/p_{n+1}(x)$ changes from positive to negative both as $x$ crosses $b$ from left to right and as $x$ crosses $c$ from left to right. Thus $p_n(x)/p_{n+1}(x)$ is negative for $x$ slightly to the right of $b$ and is positive for $x$ slightly to the left of $c$. But $p_{n+1}$ doesn't change sign on $(b,c)$, hence $p_n$ must change sign on $(b,c)$, contradiction, since $b,c$ are consecutive roots of $p_n$.

Hence $p_n$ has at most one root strictly between any two consecutive roots of $p_{n+1}$.

Since $p_n$ has $n$ distinct roots, each of which lies in one of the $n$ intervals $(y_i,y_{i+1})$, and each interval $(y_i,y_{i+1})$ contains at most one root of $p_n$, it follows by the pigeon hole principle that each interval $(y_i,y_{i+1})$ contains exactly one root of $p_n$.

Edit:

Looking back at your question, I see that your idea was to approach the problem by using the fact (assuming $p_n$ and $p_{n+1}$ have no common zeros) that for each $y_i\in \{y_1,...,y_{n+1}\}$, we have $$ \left\lbrace \begin{align*} \lim_{x\to y_i^{\Large{-}}}p_n(x)/p_{n+1}(x)&\in\{\pm\infty\}\\[4pt] \lim_{x\to y_i^{\Large{+}}}p_n(x)/p_{n+1}(x)&\in\{\pm\infty\}\\[4pt] \end{align*} \right. $$ Here's an argument based on that approach . . .

Consider the interval $(y_i,y_{i+1})$.

We know $$ \left\lbrace \begin{align*} \lim_{x\to y_i^{\Large{+}}}p_n(x)/p_{n+1}(x)&\in\{\pm\infty\}\\[4pt] \lim_{x\to y_{i+1}^{\Large{\;\;-}}}p_n(x)/p_{n+1}(x)&\in\{\pm\infty\}\\[4pt] \end{align*} \right. $$ but since $p_n(x)/p_{n+1}(x)$ is decreasing on $(y_i,y_{i+1})$, we must have $$ \left\lbrace \begin{align*} \lim_{x\to y_i^{\Large{+}}}p_n(x)/p_{n+1}(x)=+\infty\}\\[4pt] \lim_{x\to y_{i+1}^{\Large{\;\;-}}}p_n(x)/p_{n+1}(x)=-\infty\}\\[4pt] \end{align*} \right. $$ hence by the Intermediate Value Theorem, $p_n$ has at least one root in $(y_i,y_{i+1})$.

Since $p_n$ has exactly $n$ roots, and at least one root in each of the $n$ intervals of the form $(y_i,y_{i+1})$, it follows that each of the intervals $(y_i,y_{i+1})$ contains exactly one root of $p_n$.