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I can't solve this problem:

Suppose $f$ is polynomial function of even degree $n$ with always $f\geq0$.

Prove that $f+f'+f''+\cdots+f^{(n)}\geq 0$.

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2 Answers 2

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Let $g=f+f'+\cdots+f^{(n)}$ and let $h(x)=e^{-x}g(x)$. Note that $f^{(n+1)}=0$, so $$h'(x)=e^{-x}(g'(x)-g(x))=-e^{-x}f(x)\le 0,$$ i.e. $h$ is decrearing on $\mathbb R$. Since $g$ is a polynomial, $\lim\limits_{x\to+\infty}h(x)=0$. It follows that $h\ge 0$, and hence $g\ge 0$.

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  • $\begingroup$ Why lim h(x) goes to zero? $\endgroup$
    – user75086
    Jun 8, 2013 at 11:24
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    $\begingroup$ @user75086: As $x\to +\infty$, the exponential function $e^x$ tends to $\infty$ much faster than any polynomial function. $\endgroup$
    – 23rd
    Jun 8, 2013 at 11:26
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Alternatively, by a standard Maximum Value Theorem argument, the even-degree polynomial $g=f+f'+\dots+f^{(n)}$ has a global minimum at some point $a$. Then $g(a)=f(a)+g'(a)\ge 0$, so $g\ge g(a)\ge 0$.

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    $\begingroup$ It's nice and more natural than my argument. +1. $\endgroup$
    – 23rd
    Jun 8, 2013 at 11:19
  • $\begingroup$ why $g(a)=f(a)+g'(a)\ge 0$? $\endgroup$
    – user75086
    Jun 8, 2013 at 11:55
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    $\begingroup$ @user75086: Because $g'(a)=0$ and $f(a)\ge 0$. $\endgroup$ Jun 8, 2013 at 12:16

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