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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ an infinitely many times differentiable function and $f(\frac{1}{n})=0$ for each $n \in \mathbb{N}$. Show that $f^{(n)}(0)=0$ for $n=0,1,2, \dots$

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Could you give me some hint what I could do?? I got stuck right now..

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A straightforward approach using Rolle's Theorem:

Let $n\in\mathbb{N}^*$. Since $f$ is differentiable on $\left(\dfrac1{n+1},\dfrac1n\right)$ and continuous on $\left[\dfrac1{n+1},\dfrac1n\right]$, there exists $c_n\in\left(\dfrac1{n+1},\dfrac1n\right)$ such that $f'(c_n)=0$.

This shows that there exists a sequence $(c_n)_{n\in\mathbb{N}^*}$ that is decreasing and $\lim\limits_{n\to+\infty}c_n=0$ such that for all $n\in\mathbb{N}^*$, $f'(c_n)=0$. Since $f'$ is continuous at $0$, we must have: $$f'(0)=\lim_{n\to+\infty}f'(c_n)=0.$$

Now repeat the process.

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    $\begingroup$ interesting, here is a thought: can you construct such function which is not identically 0 on a region near the origin? $\endgroup$ – Lost1 Jun 29 '14 at 16:54
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    $\begingroup$ @Lost1 sure: take $$f(x)=\begin{cases}\sin\left(\dfrac\pi x\right)\exp\left(-\dfrac1{x^2}\right)&\text{if $x\neq0$}\\0&\text{if $x=0$}.\end{cases}$$ $\endgroup$ – gniourf_gniourf Jun 29 '14 at 16:57
  • $\begingroup$ I see (I remember something says you cannot make it analytic though, is it right? something like identity theorem? or am I confused?) $\endgroup$ – Lost1 Jun 29 '14 at 17:00
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    $\begingroup$ @Lost1 You're right, it's not analytic; it's only of class $C^\infty$. If a function is analytic and its zeros have an accumulation point, then it's identically nil. $\endgroup$ – gniourf_gniourf Jun 29 '14 at 17:01
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    $\begingroup$ @Lost1 An immediate corollary to the OP's question is that any analytic function satisfying $f(1/n)=0$ for all $n$ is identically zero. This is because an analytic function can be expressed in terms of its taylor series: $$f(x) = \sum_{n=1}^\infty \frac{f^{(n)}(0)}{n!} x^n = 0.$$ However, there are many $C^\infty$ functions that are not analytic. $\endgroup$ – Joel Jul 1 '14 at 16:20
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Assume $f$ is differentiable and there are $a_n \rightarrow 0$ such that $f(a_n)=0$ then appying the mean value to $[a_n, a_{n+1}]$ we get $b_n \in [a_n, a_{n+1}]$

$$f(a_{n+1})-f(a_n)=f^{\prime}(b_n)(a_{n+1}-a_n)$$ so $f^{\prime}(b_n)=0$. This should solve the problem.

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I would stick with gniourf_gniourf's solution, but a neat alternative is to use the fact that $$\lim_{h\to0} \sum_{i=0}^n {n \choose i} (-1)^{n-i} f(ih) = f^{(n)}(0)$$ now take $h = 1/m!$ to get $i/m! = 1/( 1 \cdot 2 \cdots (i-1) \cdot (i+1) \cdots m)$ in the arguments of $f$. In particular $f$ vanishes at $i/m!$.

Thus $f^{(n)}(0)$ is the limit of the zero sequence, and is zero itself.

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