$p_n(x)=p_{n-1}(x)+p_{n-1}^{\prime}(x)$, then all the roots of $p_k(x)$ are real

$p_0(x)=a_mx^m+a_{m-1}x^{m-1}+\dotsb+a_1x+a_0(a_m,\dotsc,a_1,a_0\in\Bbb R)$ is a polynomial, and

$$p_n(x)=p_{n-1}(x)+p_{n-1}^{\prime}(x),\qquad n=1,2,\dotsc$$

then, there exist $N\in\Bbb N$, such that $\forall n \geq N$, all the roots of $p_n(x)$ are real


I think, let $f_n(x)=e^xp_n(x)$, then

$$f_n(x)=\frac {d(f_{n-1}(x))}{dx} $$

we get that

$$f_n(x)=\frac{d^n\left( e^xp_0(x)\right)}{dx^n}$$

so, If there exists a $p_N(x)$ such that $p_N(x)$ has $m$ real roots, then $\forall n\geq N$, $p_n(x)$ has $m$ real roots.

I’d be grateful for any help

Solutions Collecting From Web of "$p_n(x)=p_{n-1}(x)+p_{n-1}^{\prime}(x)$, then all the roots of $p_k(x)$ are real"

It is true that, for $n$ large enough, $p_n(x)$ will have distinct real roots, which I will prove by induction on $m$.

Before proceding with the proof, let me first state what it shows about the behaviour of $p_n$ as $n$ increases. Assuming the leading coefficient is positive, $p_n^{(m-2)}$ are quadratics with a unique minimum. As $n$ increases, the minimum decreases monotonically until it becomes negative, at which point $p_n^{(m-2)}$ has distinct real roots. Then, $p_n^{(m-3)}$ is a cubic with a local maximum and a local minimum. As $n$ increases further, the local maximum increases until it is positive and the local minimum decreases until it is negative, at which point $p_n^{(m-3)}$ has distinct real roots. More generally, for each $k=0,1,2,\ldots,m-2$, once $p_n^{(k+1)}$ has distinct real roots then $p_n^{(k)}$ has $m-k-1$ local extrema. Further increasing $n$, the local maxima increase monotonically until they are positive and the local minima decrease until they are negative, at which point $p_n^{(k)}$ has distinct real roots. By induction then, $p_n$ eventually has distinct real roots.

Now, to continue with the proof.

For $m=1$, $p_n(x)$ is of first order so always has a real root.

For $m > 1$, we now suppose that the statement holds for polynomials of degree $m-1$. Then, setting $q_n(x)=p_n^\prime(x)$, we have $q_n(x)=q_{n-1}(x)+q_{n-1}^\prime(x)$. By the induction hypothesis, $q_n(x)$ has $m-1$ distinct real roots for all large enough $n$. I’ll denote these by $a_{n,1} > a_{n,2} > \cdots > a_{n,m-1}$. These are the turning points of $p_n(x)$. I’ll show that the for large $n$, the signs of $p_n(a_{n,1}),p_n(a_{n,2}),p_n(a_{n,3}),\ldots$ are alternating, from which it follows that $p_n$ has $m$ distinct roots.

By scaling, we assume wlog that $p_n(x)$ has leading coefficient $1$ (i.e., it is monic). Then, $q_n$ is positive on $(a_{n,1},\infty)$, negative on $(a_{n,2},a_{n,1})$, positive on $(a_{n,3},a_{n,2}))$, etc.
Note that $a_{n,i}$ is a local maximum of $(-1)^ip_n(x)$ (for each $i=1,2,\ldots,m-1$). This means that
$$
(-1)^ip_n(a_{n,i}+h)=(-1)^ip_n(a_{n,i})-ch^{2r}+O(h^{2r+1})
$$
for some positive $c$ and positive integer $r$. So,
$$
(-1)^ip_{n+1}^\prime(a_{n,i}+h)=-c(2r)(2r-1)h^{2(r-1)}+O(h^{2r-1}).
$$
This shows that $(-1)^ip_{n+1}(x)$ is decreasing in a neighbourhood of $a_{n,i}$.

Now, for $i=1,2,\ldots,m-2$, we see that $(-1)^ip_{n+1}(x)$ is increasing at $a_{n,i+1}$ and decreasing at $a_{n,i}$. Hence, it has a local maximum in $(a_{n,i+1},a_{n,i})$, which we will denote by $b_{i}$, so $a_{n,i+1} < b_i < a_{n,i}$ and $(-1)^ip_{n+1}(b_i) > (-1)^ip_{n+1}(a_{n-i})$. Similarly, $(-1)^{m-1}p_{n+1}(x)$ is decreasing at $a_{n,m-1}$ and (as $p_{n+1}$ is monic of degree $m$), it tends to $-\infty$ as $x\to-\infty$. Hence, it has a local maximum, $b_{m-1}$ in the range $(-\infty,a_{n,m-1})$. So, $b_{m-1} < a_{n,m-1}$ and $(-1)^ip_{n+1}(b_{m-1}) > (-1)^ip_{n+1}(a_{n,m-1})$. As $b_1 > b_2 > \cdots > b_{m-1}$ are roots of $p_{n+1}^\prime(x)$, we have $a_{n+1,i}=b_i$.

So far, we have shown that
$$
a_{n,1} > a_{n+1,1} > a_{n,2} > a_{n+1,2} > a_{n,3} > \cdots > a_{n,m-1} > a_{n+1,m-1}
$$
and, for each $i$, $(-1)^ip(a_{n,i})$ is strictly increasing in $n$. If, for each fixed $i$, it can be shown that there is an $\epsilon > 0$ with $(-1)^ip_{n+1}(a_{n+1,i})\ge(-1)^ip_n(a_{n,i})+\epsilon$ for all large $n$ then, for $n$ large enough, $(-1)^ip_n(a_{n,i})$ will be positive. So, the signs of $p_n(a_{n,1}),p_n(a_{n,2}),\ldots$ will be alternating in sign and we are done. Let us proceed by contradiction and assume to the contrary that $(-1)^ip_{n+1}(a_{n+1,i})< (-1)^ip_n(a_{n,i})+\epsilon$ (for small enough $\epsilon$ this will give a contradiction regardless of $n$).

If $i < m-1$ then $(-1)^ip_{n+1}(x)\le(-1)^ip_{n+1}(a_{n+1,i})$ on $(a_{n,i+1},a_{n,i})$. Setting $a_{n,m}=-\infty$ then this also holds for $i=m-1$. By the assumption, $(-1)^ip_{n+1}(x)\le(-1)^ip_{n}(a_{n,i})+\epsilon$ in this range so, setting $f(x)=(-1)^i(p_n(a_{n,i})-p_{n}(x))\ge0$,
$$
f(x)+f^\prime(x)=(-1)^i(p_n(a_{n,i})-p_{n+1}(x))\ge-\epsilon.
$$
Therefore,
$$
f(x)=-\int_x^{a_{n,i}}\frac{d}{dy}(e^{y-x}f(y))dy
=-\int_x^{a_{n,i}}e^{y-x}(f(y)+f^\prime(y))dy\le\epsilon\int_x^{a_{n,i}}e^{y-x}dy=\epsilon(e^{a_{n,i}-x}-1).
$$
This shows that in the range $(a_{n,i+1},a_{n,i})$,
\begin{align}
\lvert p_{n}(x)-p_n(a_{n,i})\rvert\le\epsilon(e^{a_{n,i}-x}-1).&&{\rm(1)}
\end{align}
So, $p_n$ is almost constant here for $\epsilon$ small. I’ll need the following lemma, which I’ll prove in a moment.

Lemma: Let $p$ be a monic polynominal of degree $m\ge1$. Then, for each $a < b$ we have
$$
sup_{x\in(a,b)}\lvert p(x)-p(b)\rvert\ge L(b-a)^m.
$$
where $L$ is a constant depending only on $m$.

We can apply this to (1). For the case $i=m-1$ and any $K>0$, apply to the range $[a_{n,m-1}-K,a_{n,m-1}]$ to get
$$
\epsilon(e^K-1)\ge L K^{m}
$$
which gives a contradiction if $\epsilon$ is small enough.
On the other hand, for $i< m-1$, $(-1)^i(p_n(a_{n,i})-p_n(a_{n,i+1})$ is positive and increasing in $n$ for all large enough $n$, so is greater than some fixed $\delta$. Applying (1),
$$
\delta\le (-1)^i(p_{n,i}(a_{n,i})-p_{n,i}(a_{n,i+1}))\le\epsilon(e^{a_{n,i}-a_{n,i+1}}-1).
$$
Assuming that $\epsilon \le 1$, we choose $K > 0$ such that $\delta\ge e^K-1$. Then, $K\le a_{n,i}-a_{n,i+1}$ and we can apply the lemma to the range $[a_{n,i}-K,a_{n,i}]$ to get
$$
\epsilon(e^K-1)\ge LK^m
$$
again, giving a contradiction for small $\epsilon$. QED


I’ll now prove the lemma. For any $\epsilon\in\mathbb{R}$ define the linear operator $T_\epsilon$ of the functions $f\colon\mathbb{R}\to\mathbb{R}$ by $T_\epsilon f(x)=f(x+\epsilon)$. If $f$ is a polynomial of degree $m$ with leading coefficient $c$, then $T_\epsilon f-f$ is a polynomial of degree $m-1$ with leading coefficient $mc\epsilon$. Hence, $(T_\epsilon-1)^mf(x)=m!c\epsilon^m$. On the other hand $\lVert T_\epsilon\rVert=1$ and $(T_\epsilon-1)^mf(x)$ only depends on the values of $f(y)$ with $y$ in the range $[x,x+m\epsilon]$. So, for a monic polynomial $f$ of degree $m$,
$$
m!\epsilon^m=(T_\epsilon-1)^mf(x)\le 2^m\sup_{y\in[x,x+m\epsilon]}\lvert f(y)\rvert
$$
If $p$ is a monic polynomial of degree $m$ and $a < b$, then taking $f(y)=p(y)-p(b)$, $x=a$ and $\epsilon=(b-a)/m$ gives the lemma with $L=m!(2m)^{-m}$.

Another approach.

I will prove the result step by step.

  1. $$\sum_{j=0}^m (-1)^j \binom{m}{j}j(j-1)\cdots(j-k+1)=0$$ if $0\leqslant k\leqslant m-1$, and $m!$ if $k=m$.

Proof.

Use $j\binom{m}{j}=m\binom{m-1}{j-1}$.

2-$$\sum_{j=0}^m (-1)^j \binom{m}{j}j^k=0$$ if $0\leqslant k\leqslant m-1$, and $m!$ if $k=m$.

Proof.

Follows from Step $1$ by linearity.

3-$$\sigma_k(1,2,\ldots,j-1)$$ is polynomial of degree $2k$ on $j$ with leading coefficient $\dfrac{1}{2^k k!}$.

Proof.

This is shown by induction using Newton’s identities

4-$$\sum_{j=0}^m (-1)^j\binom{m}{j}\sigma_k(1,2,\ldots,j-1)=0$$ if $2k<m$,
and $1\times 3\times\cdots (m-1)$ if $2k=m$.

Follows from Step $2$ and $3$

5 $$\sum_{j=0}^m (-1)^j \binom{m}{j}(1-\frac{1}{n})\cdots (1-\frac{j-1}{n})$$ is equivalent
when $n\rightarrow +\infty $ to $(-n)^{-m/2}\times 1\times 3\times\cdots (m-1)$ if $m$ is even and, et is $o(n^{-m/2})$ if $m$ is odd.

Proof.

Just develop the product and use step $4$.

6- For $p_0(x)=x^m$, $p_n(-n)$ is equivalent to $(-n)^{m/2}\times 1 \times
3\times\cdots\times (m-1)$ if $m$ is even, and $p_n(-n)=o(n^{m/2})$ if $m$ is odd.

Proof.

Use $p_n(x)=\sum_{j=0}^m \binom{n}{j}m(m-1)\cdots (m-j+1)x^{m-j}=\binom{m}{j}n(n-1)\cdots (n-j+1)x^{m-j}$ and substitute $x=-n$ to return to step $5$.

7-Denote $q_{m,n}(x)=n^{-m/2}p_{m,n}(-n+x\sqrt{n})$ with $p_0=x^m$
We show by induction on $n$ as $q_ {m, n} $ converges to the Hermite polynomial $H_m $ when $n$ tends to infinity.

For $n=0$ is obvious.

Suppose the assertion true at rank <$ n $. Then she is true for $\frac{1}{m}p’_{m,0}(x)=x^{m-1}=p_{m-1,0}$.

As the derivation commutes with the transformation $p_0\mapsto p_n$, we have $p’_{m,n}=mp_{m-1,n}$.

It comes $q’_{m,n}=mq_{m-1,n}$, thus $q’_{m,n}$ converges to $mH_{m-1}=H’_m$ as $m$ tends to $+\infty$.

Using step $6$ we have $q_{m,n}(0)$ converges to $H_m(0)$.

8- The result of step 7 is true for any monic polynomial of degree $m$ instead of $p_0$.

Finally, We boils down to $p_0$ unitary. By continuity of the roots in the coefficients of a polynomial and the fact that the Hermite polynomials are divided to simple roots over $\mathbb{R}$, $p_n$ is also for $n$ large enough after step $8$.

Not a solution at all, so (down)vote accordingly. Just writing down whatever my toying with Mathematica suggests in the hope that it may inspire somebody else. I think this line of attack should lead to a solution, but it does require careful estimates. Hopefully a more clever method is out there.

The formula in Abiessu’s comment shows that the transformation $T_n:p_0\mapsto p_n$ is linear
(this could also be proven directly). A more careful look at that formula reveals that for large $n$ the coefficients coming from the leading term, i.e. $T_n(a_mx^m)$, will dominate the other terms. In a way (that needs to be made more precise) this suggests that the zeros of $T_n(p)$ are close to the zeros of
$$T_n(x^m)=x^m+m{n\choose 1}x^{m-1}+m(m-1){n\choose 2}x^{m-2}+\cdots$$

So I propose the following attack:

  • Show that for large enough $n$ the polynomial $G_n:=T_n(x^m)$ has $m$ distinct zeros, and that the derivative $G_n’$ has a large enough absolute value at these zeros.
  • Show that the difference $|a_m G_n(x)-p_n(x)|$ is small enough (in comparison to $G_n'(x)$) near the zeros of $G_n$ so that we can conclude $p_n$ to also have a zero in the vicinity.

Testing shows that the zeros of $G_n$ will be “near” $x=-n$ for a suitable definition of “near”. This is hardly a surprise as it follows from the definitions that
$$
\lim_{n\to\infty}\frac{G_n(nx)}{n^m}=(x+1)^m.
$$
I checked what happens with
$$
p_0(x)=(x^2+1)(x^2+4)=x^4+5x^2+4.
$$
Since I went up to $n=50$ (way overkill, the condition is fulfilled much earlier) I will share a few plots with you.
Here are the plots of $p_{50}$ and $G_{50}$. You see that at this resolution they nearly coincide.

enter image description here

Here’s a close up of the neighborhood of one of the zeros.

enter image description here


Edit: I think I figured out a useful scaling. Consider the polynomial (changing the notation a bit, as I need double indexing)
$$
F_{m,n}=T_n(x^m):=\sum_{i=0}^n\frac{m! n! i!}{(m-i)! (n-i)!}x^{m-i}.
$$
Let’s do the translation that John Mangual also studied, and scale the variable by $\sqrt n$, and define
$$
H_{m,n}:=\frac1{n^{m/2}}F_{m,n}(\sqrt n x-n).
$$
I’m well on my way to proving (not there yet, but I have computer justification for $m\le5$, and some useful looking identities) that
$$
\lim_{n\to\infty}H_{m,n}(x)=He_m(x),
$$
where $He_m(x)$ is the so called probabilist’s Hermite polynomial of degree $m$. I would be surprised if this were not either known or easily deduced from known facts.

Anyway, I conjecture that the zeros $z_{i,n}, i=1,2,\ldots,m$, of $p_n(x)$ can be numbered in such a way that the limits
$$
z_i=\lim_{n\to\infty}\frac{z_{i,n}+n}{\sqrt n}
$$
exist, and are the zeros of $He_m(x)$.

Let’s recall two observations in this discussion:

  • $\displaystyle f_n(x)=\frac{d^n\left( e^xp_0(x)\right)}{dx^n}$

  • $\displaystyle p_n(x)=\sum_{i=0}^n{n\choose i}p_0^{(i)}(x)$

Notice also, the degree never goes up $\deg p_n = \deg p_0 = m$ for all $n \geq 0$. Then

$$ \sum_{k=0}^n{n\choose k}p_0^{(k)}(x) = \sum_{k=0}^m{n\choose k}p_0^{(k)}(x)
\approx \sum_{k=0}^m \frac{n^k}{k!}p_0^{(k)}(x) = p_0(x+n)
$$

where $m << n$. In this regime, we estimate the binomial coefficient by:
$ \binom{n}{k} \approx \frac{n^k}{k!}$ and then use Taylor formula.


Notice this only holds asymptotically for $n >> 1$ we get $p_n(x) \approx p_0(x+n)$. Then I wasn’t agreeing with Clin’s conjecture since $p_0(x)$ may not always have real roots, e.g. $p_0(x) = x^2 + 1$. In fact $p_n(x) = (x+n)^2 – n + 1$.


A second-order estimate gives $p_n(x) \approx p_0(x+n) – n \,p”_0(x+n)$…

Does $p_0(x) – n \, p_0”(x)$ have all real roots for $n$ sufficiently large?

Generically we can write $p_0(x) = x^m + ax^{m-2} + \dots$ so that
$p_n(x-n) \approx x^m – n \times m(m-1)x^{m-2} + \dots$