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Let $W\left( x\right) \ge 0$ for $x \in \mathbb{R}$ be a polynomial. Prove $$u\left( x\right)=W\left( x\right)+W'\left( x\right)+W''\left( x\right)+ \cdots \ge 0.$$ Is there a simple way?

Zhanxiong
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piteer
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2 Answers2

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. Oh, let $g(x)=W(x)+W'(x)+...$, then $g'(x)=W'(x)+W''(x)+...$ therefore $g'(x)=g(x)-W(x)$. Put $y=g(x)$, then we obtain the equation: $$ y'=y-W(x) $$ This can be solved as follows: \begin{align} & y'=y-W(x) \\ & \implies e^{-x}y' = ye^{-x} - We^{-x} \\ & \implies e^{-x}y' -ye^{-x} = -We^{-x} \\ & \implies \frac{d}{dx}(ye^{-x}) = -We^{-x} \\ \end{align}

From here, the comment below will guide you, because I cannot integrate unless I know some bounds, which are assumed in the comment below.

Sarvesh Ravichandran Iyer
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  • @Winther thanks, it's a small mistake. – Sarvesh Ravichandran Iyer Mar 29 '16 at 13:04
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    It's small but it requires changes to the final argument. You can also not take $c=0$ unless you know a $a$ where $u(a) = 0$. Integrating gives us $u(x) = e^{x-a}\left[u(a) + \int_x^a W(t)e^{a-t}{\rm d}t\right]$. You can use that $u(a)$ is positive for large enough $a$ since $u$ is off even degree. Then $W\geq 0$ implies $u(x)\geq 0$ for $x\leq a$ and since $a$ is arbitrary the result follows. – Winther Mar 29 '16 at 13:21
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Suppose for a contradiction $u(x_0)<0$ for some point $x_0\in \mathbb{R}$. Then, because $u$ is a polynomial of even degree with positive leading coefficient (because $W$ must have these properties as well), there must be an interval $[a,b]\ni x_0$ such that $u(x)<0$ for all $x\in ]a,b[$ and $u(a)=0=u(b)$. Then there must be some $c\in ]a,b[$ such that $u'(c)=0$ by Rolle's theorem. Hence $$0=u'(c)=W'(c)+W''(c)+\dots =u(c)-W(c),$$and thus $$0\leq W(c)=u(c)<0,$$ which is a contradiction.

Arnaud D.
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