Iterated antiderivatives of polynomials having many real roots

Question For which polynomials $$p_n:\mathbb{R} \rightarrow \mathbb{R}$$ having $$n$$ distinct real roots can we find an infinite sequence of polynomials $$p_n, p_{n+1}, p_{n+2} , p_{n+3}, \dots,$$ such that $$p_i$$ is a polynomial of degree $$i$$ with $$i$$ distinct real roots and $$p_{i+1}$$ is an anti-derivative of $$p_i$$ for all $$i \geq n$$?

Hermite polynomials have this property. The $$n-$$th Hermite polynomial $$H_n$$ has $$n$$ distinct roots and satisfies the derivative relation $$\frac{d}{dx} H_n = n H_{n-1}.$$ This means that when computing the antiderivative of $$H_{n}$$, one can always find the correct constant to turn the anti-derivative into a multiple of $$H_{n+1}$$ and multiples of $$H_{n+1}$$ have exactly $$n+1$$ roots. My gut feeling would be that this is a very special property that not many polynomials have and I am wondering whether there is any sort of result in that direction.

Motivation This question is motivated by the asymptotic behavior of roots of polynomials when polynomials are differentiated many times. A while back I proposed a PDE to describe this and the PDE seems to be smoothing, so there should be a loss of information as one differentiates. More recently, Jeremy Hoskins and I proved a result that also hints towards loss of information. This sort of suggests that having these infinite chains should be somehow `rare' in some sense but it's more a vague connection than anything theorem-based.

• The discriminant of $p$ is polynomial in the coefficients of $p$ and vanishes just when $p$ has multiple roots. So given $p$, the discriminant of $p + C$ is a polynomial in $C$, and so only finitely many choices of $C$ will have $p + C$ have multiple roots. So you should always be able to do this. – Kevin Casto Apr 4 at 3:24
• I don't think it's always possible. If we take (-1 + x) (-0.8 + x) (0.8 + x) (1 + x), then no matter what constant we choose for the antiderivative, the antiderivative cannot have more than 3 real roots. The problem is not only about the roots being distinct, they should also all be real. – Stefan Steinerberger Apr 4 at 4:05
• The question is equivalent to asking for which power series $F(x)\in\mathbb{C}[[x]]$ is it true that if $e^{tx}F(x)=\sum_{n\geq 0}P_n(t)\frac{x^n}{n!}$, then all the roots (or zeros) of $P_n(t)$ are real. This seems to be quite a rare property. Possible examples other than $e^{-x^2}$ and some related ones like $(1+x)e^{-x^2}$ are $\cosh\sqrt{x}$ and $\sqrt{x}\sinh\sqrt{x}$. – Richard Stanley Apr 6 at 0:38
• Even more generally, power series like $F(x)=\sum_{n\geq 0}\frac{x^n}{n!(2n+1)!(5n+3)!}$ also seem to work (if Maple can be trusted). – Richard Stanley Apr 6 at 1:50
• @StefanSteinerberger, for the thrill of receiving my first bounty, I will convert my comment to an answer. – Richard Stanley Apr 7 at 1:35

1 Answer

The question seems to be answered at the "Untitled" link from canvas.wisc.edu when you do a google search on "appell sequence real zeros".