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Given a function defined by $a_n(x) = \frac{d^n}{dx^n} e^{-x^2}$. Every function can be written as $a_n(x) = h_n(x) e^{-x^2}$ where $h_n(x)$ is a polynomial (the Hermite-Chebyshev polynomial of degree $n$ to be precise). I want to prove that every polynomial $h_n(x)$ has $n$ real valued roots.

I am trying to prove this by induction. Since $a_0(x)$ is symmetric and has no roots, $a_1(x)$ has one root (Rolle's theorem).

Now given $a_n(x)$ has $n$ roots. From Rolle's theorem I get that $a_{n+1}(x)$ must have at least $n-1$ roots. But there are two more. $a_{n+1}(x)$ has a root between $-\infty$ and the first root of $a_n(x)$ and between at the last root of $a_n(x)$ and $+\infty$.

I know that $\lim_{x \to \pm \infty} a_n(x) = 0$. So my question is: Is Rolle's theorem valid for limits?

When there is a $f(x_0) = \lim_{x \to \infty} f(x)$, does it necessarily follow that $f'(x) = 0$ somewhere in $[x_0, \infty)$? Can you prove this?

Thomas Andrews
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1 Answers1

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If $f$ is a differentiable function on the real line, and $f(x)\to 0$ as $x\to+\infty$ and $f(a)=0$ for some $a\in\mathbb R$ then $f'(b)=0$ for some $b\in(a,+\infty)$.

Proof: Show that $f(x)$ must have a maximum or minimum on $(a,+\infty)$.

Alternatively, let $g(x)= f(a-1+ 1/x)$ for $x\in(0,1]$ and $g(0)=0$. Then $g(x)$ is a continuous function on $[0,1]$ and $g(0)=g(1)=0$. But $g'(x)=\frac{-1}{x^2}f'(a-1+1/x)$ for $x\in(0,1)$, so we can apply Rolle's theorem to show there must be a $b\in(0,1)$ such that $g'(b)=0$. But then we conclude that $f'(a-1+1/b)=0$.

Thomas Andrews
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