6
$\begingroup$

$p(x)\geq 0 \forall x \in \mathbb{R} \Rightarrow p(x)+p'(x)+p''(x)+...+p^{(n)}(x)\geq 0$, where p(x) is a polynomial of degree n.

I showed:

$a_{n}+...+a_{0}\geq 0$,

$p(x)+p'(x)+p''(x)+...+p^{(n)}(x)=\sum_{m=0}^{n}\sum_{k=m}^{n}a_{k}\frac{n!}{(n-k)!}p(x)^{(k-m)}=\sum_{k=0}^{n}\sum_{m=0}^{k}a_{k}\frac{n!}{(n-k)!}p(x)^{(k-m)}$

$\endgroup$
1
  • $\begingroup$ Interesting remark : The degree of $p$ must be even. $\endgroup$ Feb 9, 2014 at 22:06

3 Answers 3

8
$\begingroup$

If $n$ is odd then $p(x)$ becomes negative as $x\to\infty$ or $x\to-\infty$. Hence we may assume $n$ is even. The function $f(x)= p(x)+\ldots +p^{(n)}(x)$ is also a polynomial of degree $n$ and hence assumes its global minimum at some $a\in\mathbb R$. Then $f'(a)=0$. But from $p^{(n+1)}(x)=0$ we see $f'(x)=f(x)-p(x)$ and hence $f(a)=p(a)\ge 0$. As $a$ is the global minimum, $f(x)\ge f(a)\ge 0$ for all $x$.

$\endgroup$
2
  • 1
    $\begingroup$ I had found it and two seconds later you posted your answer... gnahhh. Good job, +1 $\endgroup$ Feb 9, 2014 at 22:08
  • $\begingroup$ The same answer was posted two years ago by user20266. $\endgroup$
    – TonyK
    Feb 10, 2014 at 0:40
4
$\begingroup$

An another proof,

$Q=P+P'+...+P^{(n)}$

Notice $$ P=Q-Q' $$ Thus $$ \forall x\in \mathbb{R}, Q(x)\geq Q'(x). $$ Let $$ \forall x \in \mathbb{R}, \varphi(x)= e^{-x}Q(x) $$ So, $$ \varphi'(x)=e^{-x}(Q'(x)-Q(x))\leq 0 $$ Therefore $\varphi$ is decreasing on $\mathbb{R}$. She tends to $0$ when $x \longrightarrow +\infty$, so $Q(x)\geq 0$

$\endgroup$
3
$\begingroup$

Notice

$$\sum_{k=0}^n p^{(k)}(x) = \sum_{k=0}^n p^{(k)}(x)\left(\int_0^\infty \frac{t^n}{n!} e^{-t}dt \right) = \int_0^\infty \left(\sum_{k=0}^n \frac{p^{(k)}(x)}{n!}t^k\right) e^{-t}dt\\ = \int_0^\infty p(x+t) e^{-t} dt $$ If $p(x) \ge 0$ for all $x$, then $\sum\limits_{k=0}^n p^{(k)}(x)$ is the integral over a non-negative function and hence is non-negative itself.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.