Let $p(x)$ be a polynomial of degree $n$ satisfying $p(x)\geq 0$ for all $x$. That is, for all $x$, $p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \geq 0$, $a_n\neq 0$.

Show that $p(x)+p'(x)+p''(x)+\cdots+p^{(n)}(x)\geq 0$ for all $x$ where $p^{(i)}(x)$ is the $i^\text{th}$ derivative.

My interest: I know that, we can rewrite the LDE as follows

$$ p(x)+p'(x)+p''(x)+\cdots+p^{(n)}(x) = Lp(x) $$

where $L := I + D + D^2 + \cdots + D^{(n)}$. Can we say anything about a linear operator of this kind so that it does not change the sign of the input it takes? I can try to solve the question by writing for all the derivatives and factoring them using $p(x)$, but I think there should be a clever way of showing this by the properties of $L$. Can I figure out a solution by just looking at $L$ and the sign of $p(x)$ as in the question? Where should I look for that?

  • $\begingroup$ I seem to remember that there was a slick trick exploiting that $(I-D)Lp = p \geq 0$, but I don't remember how that worked. $\endgroup$ – t.b. Dec 4 '11 at 14:56

Instead of $L_n := (1 + D + D^2 + \dots + D^{n})$, use $L_{\infty} := (1 + D + D^2 + \dots)$, which comes to the same thing for polynomials of degree $n$ or less. If we let

$$\sigma(x) = p(x)+p'(x)+p''(x)+\dots+p^{(n)}(x) = L_{\infty}(p)(x)$$

then we can sum the geometric progression in $D$ to get

$$\sigma(x) = ((1 + D + D^2 + \dots)p)(x) = ((1-D)^{-1}p)(x)$$


$$p(x) = ((1-D)\sigma)(x)$$

This might look like trickery, but if you check it against the original expression for $\sigma$ you can see that it works:

$$((1-D)\sigma)(x) = p(x)+p'(x)+p''(x)+\dots+p^{(n)}(x) - (p'(x)+p''(x)+\dots+p^{(n)}(x))$$ $$= p(x)$$

Now define $\tau(x) = e^{-x}\sigma(x)$, so $\tau'(x) = e^{-x}(\sigma'(x) - \sigma(x)) = e^{-x}((D-1)\sigma)(x) = -e^{-x}p(x)$. By hypothesis, this is $ \le 0$ for all $x \in \mathbb R$. Also, $\sigma$ is a polynomial, so $\tau(x) \to 0$ as $x \to \infty$. Therefore $\tau(x) \ge 0$ for all $x$.

Hence $\sigma(x) \ge 0$ for all $x$, which is what we wanted.

  • 1
    $\begingroup$ +1. A similar argument shows that $p(x)+tp'(x)+t^2p''(x)+\ldots+t^np^{(n)}(x)\gt0$ for every $x$ and every $t\geqslant0$. $\endgroup$ – Did Dec 4 '11 at 15:51
  • $\begingroup$ @Didier: Thank you for correcting my mistake. I have reverted a couple of your gratuitous edits, because it is, after all, my response. $\endgroup$ – TonyK Dec 4 '11 at 15:55
  • $\begingroup$ Of course. Sorry for the trouble. Nice answer. $\endgroup$ – Did Dec 4 '11 at 15:56
  • $\begingroup$ Very nice! Not that it matters much, but there is now a switch from $1$ to $I$ in the second displayed formula (meaning: before that you write $1$ consistently, afterwards $I$). $\endgroup$ – t.b. Dec 4 '11 at 15:59
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    $\begingroup$ @Didier: That is a question of style, not formal correctness. $\endgroup$ – TonyK Dec 4 '11 at 17:09

I have to admit the following solution was proposed to me by a friend, I did not find it:

Let $$f(x) = p(x) + p^\prime(x) + ... + p^{(n)}(x) $$

Note that $$ f^\prime = p^\prime + p^{\prime \prime} + ...+ p^{(n)} $$

that is, $$ f = p + f^\prime$$

Clearly $n$ is even. Hence $f$ has even degree, too. This implies that $f$ attains it's absolute minimum (it's not a maximum, as $f$ behaves like $p$ at infinity) in some point $z_0$, hence $ f^\prime(z_0) = 0.$ Consequently, $\forall x$, $$f(x) \ge \min(f) = f(z_0) = p(z_0) + f^\prime (z_0) = p(z_0) \ge 0 $$

  • $\begingroup$ That's nice! And it seems to be completely different from my answer. After getting $f = p + f'$, your (friend's) answer uses the fact that $f$ attains an absolute minimum; my answer uses the fact that $e^{-x}f(x) \to 0$ as $x \to \infty$. So between them, the two approaches cover more classes of function that either answer alone. $\endgroup$ – TonyK Dec 5 '11 at 20:54

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