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Let $P(x)$ be a polynomial of degree $n$ and $P(x) \geq 0$ then prove that the polynomial $f(x)=P\left(x\right)+P'\left(x\right)+...+P^{(n+1)}\left(x\right)$ has no real roots. [$P^{(k)}(x)$ means the $k^{\text{th}}$ derivative of $P(x)$]

$P(x)$ is clearly an even degree polynomial, and so is $f(x)$. Now assume it does have at least two real roots. Then by Rolle's theorem, we get some $c$ such that $f'(c)=0 \implies f(c)=P(c)$. How do I proceed from here?

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    $\begingroup$ Hint: consider the global minimum of $f(x)$. $\endgroup$
    – Greg Martin
    Commented Feb 3 at 5:45
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    $\begingroup$ [1] Let Degree=2 $P(x)=Ax^2+Bx+C$ where $A$ is Positive , then $f(x)=Ax^2+Bx+C+2Ax+B+2A$ : Check why this has no real roots. [2] Check with Degree=4. [3] We can then generalize it to all Even Degree. [4] Basically we have to show that $f(x)$ is "Controlled" by Positive $A$ here. $\endgroup$
    – Prem
    Commented Feb 3 at 5:52
  • $\begingroup$ Hint $:$ $f'(x) - f(x) = - P(x).$ $\endgroup$
    – Anacardium
    Commented Feb 3 at 6:03
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    $\begingroup$ Oh , BTW , your Solution is "almost" complete , with the hint by @GregMartin , where you got $P(c)$ is the minimum of $f(x)$ : We know that $P(c)$ is Positive , hence there is no Zero-Crossing for $f(x)$ , hence $f(x)$ can have no real roots ! $\endgroup$
    – Prem
    Commented Feb 3 at 6:04
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    $\begingroup$ In your definition of $f$ isn't the last term $0$? $\endgroup$
    – geetha290krm
    Commented Feb 3 at 6:23

1 Answer 1

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$f=f'+P$, since $P\geq 0$, $f'-f\leq 0$ so $(fe^{-x})'\leq 0$ and $fe^{-x}$ is decreasing. But $fe^{-x}$ tends to $0$ when $x\to +\infty$, so $f$ is either identically $0$ in $[c,+\infty)$ if it has a zero $c$, or positive on the real line. But $f$ is a polynomial of degree $n>0$, so $f$ can only be positive.

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    $\begingroup$ What happens if $f$ has a zero? Why? I can't understand well without plotting. $\endgroup$
    – Bob Dobbs
    Commented Feb 3 at 10:34
  • $\begingroup$ @Bob Dobbs, $y(x) = f(x)e^{-x}$ is decreasing continuous function, so the image of segment $[-n,n]$ is $[y(n),y(-n)]$ and moreover $\lim_{x\to - \infty}y(x) = +\infty$ and $\lim_{x\to + \infty}y(x) = 0$, so the range of $y(x)$ is contained in $[0,+\infty)$. If $y$, i.e. $f$ has a zero, then by monotonicity of $y$, $y$ is zero on $[c,+\infty)$, so $f$ is zero on $[c,+\infty)$, so $f$ is zero polynomial. $\endgroup$
    – Ennar
    Commented Feb 3 at 11:01

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