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Let us say that a polynomial with real coefficients is totally real if all its complex roots are real and distinct. Let $P \in \Bbb R [X]$ be totally real. Is it true that

$$Q(X)=\int_0^XP(t)\,dt+aP(X)$$

is also totally real for all real $a$? If not, is this true if one adds the condition $P(0)=0$? Or some other additional condition?

As an additional question, can one give an upper bound on the roots of $Q$ in terms of those of $P$?


EDIT: Sorry, this is false in general, it is in fact easy to give counterexamples. My question is then: what reasonable additional condition should one add ?

Sorry if this is an elementary question.

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    $\begingroup$ You need distinct real roots, no ? $\endgroup$
    – F. C.
    Dec 3, 2020 at 10:59
  • $\begingroup$ @F.C.: thank you! corrected the post accordingly $\endgroup$ Dec 3, 2020 at 11:03
  • $\begingroup$ If we can express P as $Q'$ for some real-rooted polynomial Q, with $Q(0)=0$. Then this question is the same as asking if aQ(x)+Q(x) is always totally real... and this is the case as the roots of Q and Q' interlace. $\endgroup$ Dec 3, 2020 at 11:54
  • $\begingroup$ @Per Alexandersson: is'nt what you are saying equivalent to: if the statement is true for $a=0$ then it is true for all $a$ ? $\endgroup$ Dec 3, 2020 at 12:06
  • $\begingroup$ Since $Q'/P'=P/P'+a$ and $P'$ interlace $P$. We have $Q'$ is totally real and $P'$ interlace $Q'$, for all $a$. $\endgroup$
    – CHUAKS
    Jan 3, 2021 at 16:27

2 Answers 2

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Nope. Lets look at the case a=0.

If you have a positive quadratic polynomial, at its second root the antiderivative has a minimum. It is easy to construct an example where this minimum is positive.

As an example take $P(x)=(x-3)^2-1=x^2-6x+8$. Its antiderivative is $Q(x)=\frac{1}{3}x^3-3x^2+8x=x(\frac{1}{3}x^2-3x+8)$. It has only one real root at $x=0$, because the discriminant of the quadratic term is negative.

Adding the condition $P(0)=0$ doesn't help. Firstly, in that case the antiderivative has a double root at $x=0$, since $Q(0)=0$ by construction and $Q'(0)=P(0)=0$. So $Q$ isn't totally real.

Secondly you can constuct a $P$, for which the antiderivative has non-real roots: Let $R(x)=x(x-1)(x-2)$. $R$ has real roots at $x=0,1,2$. Its antiderivative has a minimum and double root at $x=2$. Now for some small enough $\epsilon>0$, $P(x)=R(x)+\epsilon x$ still has three distinct real roots, but the minimum of the antiderivative goes above $0$.

These constructions also hold for all $a$, that are small enough that the minimum of the antiderivative stays above 0.


EDIT: I overlooked your EDIT, sorry

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Not an answer but there are linear operators sending any totally real $p(x)$ of degree $n$ to another one with degree $n+1$ with interlacing roots. If $a,t$ are real numbers with $t>0$ and $p(x)=\prod_{j=1}^n(x-c_j)$ is totally real, and we let $q(x):=(x+a-t\frac{d}{dx})p(x)$, then $q/p$ has partial fraction $f(x):=q(x)/p(x)=x+a-\sum_{j=1}^n\frac{t}{x-c_j}$. The graph of $f$ has $n+1$ vertical monotonically increasing branches asymptotic to $y=x$ at $\pm \infty$. For any real $c$, the numerator of $f(x)-c$ defines a degree $n+1$ totally real polynomial interlacing $p$ (in particular the polynomial $q$ for $c=0$) with roots insides the intervals delimited by $c_j$. If we have real sequence $a_j,t_j>0,j=1,2,3...$, starting with $p(x)$ of any degree, one define in this way infinitely many interlacing sequences $\prod_{j=1}^m(x+a_j-t_j\frac{d}{dx})p(x),m=1,2,...$ of totally real polynomials. This construction generalizes the classical (probabilist's) Hermite polynomial which is just $H_n(x)=(x-\frac{d}{dx})^n[1]$. In particular, this prove $H_n(x)$ are totally real with interlacing roots.

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