# Prove that the polynomial $f(x)=P\left(x\right)+P'\left(x\right)+...+P^{(n+1)}\left(x\right)$ has no real roots [duplicate]

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?

• Hint: consider the global minimum of $f(x)$. Commented Feb 3 at 5:45
• [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.
– Prem
Commented Feb 3 at 5:52
• Hint $:$ $f'(x) - f(x) = - P(x).$ Commented Feb 3 at 6:03
• 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 !
– Prem
Commented Feb 3 at 6:04
• In your definition of $f$ isn't the last term $0$? Commented Feb 3 at 6:23

$$f=f'+P$$, 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.
• What happens if $f$ has a zero? Why? I can't understand well without plotting. Commented Feb 3 at 10:34
• @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. Commented Feb 3 at 11:01