number of real roots of $p(x) +p'(x)$

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 $$p(x)+p'(x)$$

• Yes, I corrected that Sep 13, 2019 at 16:13

2 Answers

For simplicity, first assume that the roots of $$p$$ are additionally distinct. Let $$x_1 be the roots of $$p$$. Then the roots $$\xi_i$$ of $$p'$$ are between these, i.e., $$x_1<\xi_1. 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.

• 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 $g'(x)$. Sep 13, 2019 at 16:56
• @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 Sep 13, 2019 at 18:30
• @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 :-) Jan 18 at 12:33

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).