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Let $P$ be a polynomial, it has $n$ distinct real roots and all these roots are larger than 1. Let $$ Q(x)=(x^2+1)P(x)P'(x)+x[(P(x))^2+(P'(x))^2]. $$
Prove $Q(x)$ has at least $2n-1$ distinct real roots.

I start to find $Q(x)=0$ between two roots. Assume $P(x)<0$ in $(x_1,x_2)$ and $P(x_1)=P(x_2)=0$. There exists a minimal value at $\xi\in(x_1,x_2)$ s.t. $$ P(\xi)<0, P'(\xi)=0, P''(\xi)>0. $$ Then I get $P(\xi)+P'(\xi)<0$.

It can be deduced by contradiction that not all $x\in (x_1,x_2)$ satisfy $P(x)+P'(x)\leq 0$. Hence there exists $\eta$ s.t. $P(\eta)+P'(\eta)>0$, then there exists $\alpha$ s.t. $$P(\alpha)+P'(\alpha)=0.$$ $Q(\alpha)<2xP(\alpha)P'(\alpha)+x[((P(\alpha))^2+(P'(\alpha))^2]\leq 0$, i.e. $$Q(\alpha)<0.$$ If I have $Q(x_1)>0$ and $Q(x_2)>0$ then by mean value theorem there are two roots in $(x_1,x_2)$.

The problems of this thought are:

(1) Even I have two roots in every interval, only $2n-2$ roots be found;

(2) $Q(x_1)>0$ and $Q(x_2)>0$ may be zero and my process would be useless?

Hope someone give me an answer or a hint.

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  • $\begingroup$ @TZakrevskiy Thank you! I've updated. $\endgroup$ – yahoo Sep 27 '17 at 12:52
  • $\begingroup$ Does your assumption implicitly imply that $P$ has degree $n$, i.e. all roots are of multiplicitly one? $\endgroup$ – MooS Sep 27 '17 at 12:54
  • $\begingroup$ @MooS I'd doubt tahat. In fact, any multiple root of $P$ is a common root of $P$ and $P'$, hence also a root of $Q$. $\endgroup$ – Hagen von Eitzen Sep 27 '17 at 12:58
  • $\begingroup$ Neither are mentioned in the book. Hence no degree and multipilicity restriction. @MooS $\endgroup$ – yahoo Sep 27 '17 at 12:59
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    $\begingroup$ if $h(x)=f(x)g(x)$ then $deg[h(x)]=deg[f(x)]+deg[g(x)]$ and if $deg[f(x)]=n $ then $deg[f'(x)]=n-1$ since f is a polynomial? Correct? does this help at all? $\endgroup$ – AmateurMathPirate Sep 27 '17 at 13:29
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Ad (1): Note that $P(x)$ and $P'(x)$ have opposite signs for $x\ll 0$, hence $Q(x)<0$ for $x\ll 0$. But at the first real root $x_0$ of $P$, we have $Q(x_0)=x_0P'(x_0)^2\ge0$ because $x_0\ge1$. Thus there is an additional root of $Q$ in $(-\infty,x_0]$.

Ad (2): If $P(x_i)=0$, then $Q(x_i)=x_iP'(x_i)^2\ge 0$. Hence if both $x_i$ are only simple roots, then we obtain two distinct root insinde $(x_1,x_2)$. Whereas if $P(x_i)$ is a $k$-fold root, $k\ge 2$, then $x_i$ is a precisely $(2k-2)$-fold root of $Q$. So first of all, the $Q(x_i)$ count as distinct roots by themselves (but that''s too few). But at least we know that $Q$ has no sign change at $x_i$. Does that help?

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I claim that $$H(x) := \frac{Q}{xP(x)P'(x)} = \frac{x^2+1}{x}+\frac{P^2+P'^2}{PP'}$$ has at least $2n-2$ roots in its range of definition. Then in particular $Q$ has at least $2n-2$ roots.

As you have proposed, we fix two roots $x_1 < x_2$ with no more roots in between them and some intermediate value $a$ with $P'(a)=0$ (Rolle's Theorem).

Let us assume $P(a) > 0$, the other case is dealt with the same way.

Then $P'<0$ in $(a,x_2)$.

Note that $|P(a)| > |P'(a)|$ but $|P(x_2)| \leq |P'(x_2)|$, hence at some point in $(a,x_2)$ we have $|P|=|P'|$. At this point we have $|\frac{P^2+P'^2}{PP'}| =2$ and thus $\frac{P^2+P'^2}{PP'}=-2$. We also know $\frac{x^2+1}{x}>2$ in $(a,x_2)$, because $x>1$.

Hence $H>0$ at this point and clearly $\lim\limits_{x \to a, x>a} H = -\infty = \lim\limits_{x \to x_2, x<x_2} H$, thus we obtain two roots of $H$ in $(a,x_2)$ by the intermediate value theorem.


Doing this for any two adjacent roots of $P$, this addds up to $2n-2$ roots, i.e $Q$ has at least $2n-2$ roots.

Here comes the punchline: $Q$ is a polynomial of odd degree, hence the number of real roots is odd. Thus having at least $2n-2$ real roots actually ensures having at least $2n-1$ roots.

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  • $\begingroup$ I agree that $\bigg|\frac{P^2+P'^2}{PP'}\bigg| =2$ , but how do you see that $\frac{P^2+P'^2}{PP'}=-2$ and not $+2$ ? $\endgroup$ – Ewan Delanoy Sep 27 '17 at 13:37
  • $\begingroup$ $P$ is positive and $P'$ is negative in $(a,x_2)$, hence the fraction is negative. $\endgroup$ – MooS Sep 27 '17 at 13:38
  • $\begingroup$ I see now, thanks. $\endgroup$ – Ewan Delanoy Sep 27 '17 at 13:40

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