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This is problem of an exam I was unable to answer.

If $p(x)$ is a polynomial of degree 2019, all roots are real. How many real roots does $p(x)+p'(x)$ have?

I know that if the polynomial $p(x)$ has $n$ real roots, then $p'(x)$ has $n-1$ real roots, but I am clueless about how to proceed with $p(x)+p'(x)$

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  • $\begingroup$ Yes, I corrected that $\endgroup$
    – J.C.VegaO
    Sep 13, 2019 at 16:13

2 Answers 2

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For simplicity, first assume that the roots of $p$ are additionally distinct. Let $x_1<x_2<\ldots <x_{2019}$ be the roots of $p$. Then the roots $\xi_i$ of $p'$ are between these, i.e., $x_1<\xi_1<x_2<\xi_2<\ldots$. Now $p(x_i)+p'(x_i)=p'(x_i)$ changes sign with every step $i$ makes because $p'$ changes sign precisely at the $\xi_i$. We conclude that $p+p'$ has a root between $x_i$ and $x_{i+1}$ for all $i$. This gives us $2018$ distinct real roots of a degree $2019$ polynomial. As non-real roots come in pairs, we conclude that all $2019$ roots of $p+p'$ must be real. (In case of interest, the additional root is before $x_1$ or after $x_{2019}$, depending on which end of $p$ tends to $-\infty$).

What if some roots of $p$ are multiple? You can add a suitable lower-degree perturbation to $p$ to separates these and conclude as above. As the perturbation can be made arbitrarily small (in our region of interest), the conclusion that all roots of $p+p'$ are real also holds in the limiting case.

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    $\begingroup$ A slight variant of this answer is to define $g(x) = p(x)e^x$ and note that $p'(x) + p(x) = 0$ is the same as $g'(x) = 0$ and then apply Rolles theorem to each adjacent pair of roots of $g$ (same as those of $p$) to find 2018 roots of $g'(x)$. $\endgroup$
    – Winther
    Sep 13, 2019 at 16:56
  • $\begingroup$ @Hagen von Eitzen Could you elaborate on this "… . Now p(xi)+p′(xi)=p′(xi) changes sign with every step i makes because p′ changes sign precisely at the ξi. We conclude that p+p′ has a root between xi and xi+1 for all i...." I am confused. This equality: p(xi)+p′(xi)=p′(xi), holds only when evaluating at the roots xi of p(x), so it tells me that f(x)= p(x)+p'(x) coincides with p'(x) only at those xi $\endgroup$
    – J.C.VegaO
    Sep 13, 2019 at 18:30
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    $\begingroup$ @J.C.VegaO if it is still of interest I have an explanation for that: $p$ doesn't change sign in $(x_i,x_{i+1})$, while $p'$ does after $\xi_i$. Because $p$ goes to $0$ in $x_2$ while $p'$ has changed sign you have that $p+p'$ does the same and so has a root. Otherwise do a graph to make it clearer :-) $\endgroup$
    – Tortar
    Jan 18 at 12:33
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Suppose $\deg P=d$ and $P$ has $d_0$ distinct roots $x_1,\ldots,x_{d_0}$ with respective multiplicity $n_i\geq 2$. Let $\Sigma = \sum_{i=1}^{d_0} n_i$ and let $x_1',\ldots,x_{d-\Sigma}'$ denote the simple roots of $P$, so that $$P(X)=\prod_{i=1}^{d_0}(X-x_i)^{n_i}\prod_{i=1}^{d-\Sigma}(X-x_i')$$

Then $P(X)+P'(X)= \prod_{i=1}^{d_0}(X-x_i)^{n_i-1}Q(X)$ for some polynomial $Q$ with $\deg Q = d-(\Sigma -d_0)=d+d_0-\Sigma$.

Since $(e^xP(x))'=e^x(P(x)+P'(x))$, Rolle's theorem applied at the $d_0+d-\Sigma$ distinct roots of $P$ yields $d_0+d-\Sigma-1$ new distinct roots of $P+P'$ (they are roots of $Q$). But since $\deg Q =d+d_0-\Sigma$, this implies that $Q$ only has real roots (one of them can be repeated once).

So $P+P'$ only has real roots. Besides, it has exactly $2d_0+d-\Sigma-1$ or $2d_0+d-\Sigma$ distinct roots (while $P$ had $d_0+d-\Sigma$ distinct roots).

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