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I have $P(x)$ a polynomial with degree $n$ ,$P(x) \ge 0$ for all $x \in$ real.

I have to prove that:

$f(x)=P(x)+P'(x)+P"(x)+......+P^{n}(x) \ge 0$ for all $x$.

I tried different methods to solve it but I got stuck.Any suggestion or advice is welcomed.

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  • $\begingroup$ Where exactly you got stuck? $\endgroup$
    – Inceptio
    Commented May 21, 2013 at 8:08
  • $\begingroup$ Inceptio I tried to explicitate every polynom P(x)=a_nx^n+a_n-1*x^n-1.....+a_1*x+a_0 P'(x)=na_nx^n-1+(n-1)a_n-1*x^n-2+....+a_1.........................................................P of order (n)=n! a_n..and I tried to sum these polynomials but I didn't get any result. $\endgroup$
    – user910
    Commented May 21, 2013 at 8:10
  • $\begingroup$ You are on a right track. $P^n(x)=n!a_n+(n-1)!$, what can you say about $P^{n-1}(x)$? You are close, give it a try. $\endgroup$
    – Inceptio
    Commented May 21, 2013 at 8:26

2 Answers 2

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you can let $$G(x)=f(x)e^{-x},-\infty<x<\infty$$ then we have $$G'(x)=f'(x)e^{-x}-f(x)e^{-x}=-P(x)e^{-x}\le 0$$ so this $G(x)$ is decreasing and $$\lim_{x\to+\infty}G(x)=\lim_{x\to+\infty}f(x)e^{-x}=0$$ $$f(x)e^{-x}=G(x)\ge \lim_{x\to+\infty}G(x)=0$$ so $$f(x)\ge 0$$

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    $\begingroup$ Thank you.I liked you solution very much.I would have never thought to use e^-x. $\endgroup$
    – user910
    Commented May 21, 2013 at 8:38
  • $\begingroup$ Hi there. Your solution looks ingenious. How did you come to think of it? I mean, what was your motivation to think of use of $e^x$ in such a nice way? $\endgroup$
    – le4m
    Commented Dec 9, 2013 at 11:33
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Because $f$ is a nonzero polynomial, it has only finitely many zeros $x_1< x_2 < \dots < x_r$. Moreover, $f(x) \underset{x \to \pm \infty}{\sim} P(x)$ so $P(x) \geq 0$ on $(- \infty,x_1]$ and $[x_r,+ \infty)$. Notice also that the sign of $f$ is constant on $[x_i,x_{i+1}]$.

Let $1 \leq i \leq r-1$. Because $f(x_i)=f(x_{i+1})=0$ there exists $y_i \in (x_i,x_{i+1})$ such that $f'(y_i)=0$ by mean value theorem. But $f(y_i)=P(y_i)+f'(y_i) \geq 0$. Therefore, $f$ is non-negative on $[x_i,x_{i+1}]$.

Finally, you deduce that $f$ is non-negative.

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