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I am looking for a function $f:\mathbb{R} \to \mathbb{R}$ that satisfies these properties:

i) $f$ is infinitely differentiable.

ii) $f$ and all its derivatives should intersect the $x$-axis only at the origin.

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If I am not mistaken, such a function does not exists.

a) We have $f(x)\not =0$ on $I=]0,+\infty[$. We may suppose that $f(x)>0$; if not, we replace $f$ by $-f$.

b) We know that $f^{(n)}(x)\not =0$ on $]0,+\infty[$. Hence it has a constant sign on $I$; the Taylor-Lagrange formula at $0$ is $\displaystyle f(x)=\frac{x^n}{n!}f^{(n)}(c)$ for some $c\in (0,x)$, and show that $f^{(n)}(x)>0$ for all $n$ and $x$ on $I$ (and all the $f^{(n)}$ are increasing on $I$).

c) We use the Taylor-Lagrange Formula for $2$ at the point $1$; we get $$f(2)=\sum_{j=0}^k \frac{f^{(j)}(1)}{j!}+\frac{f^{(k+1)}(c)}{(k+1)!}$$ for some $c\in (0,2)$. As all the terms are positive, we have $\displaystyle \frac{f^{(k)}(1)}{k!}\leq f(2)$ for all $k$.

d) Let $x\in (0,1)$. We have for a $c\in (0,x)$ $$f(x)=\frac{x^n}{n!}f^{(n)}(c)\leq \frac{x^n}{n!}f^{(n)}(1) \leq x^nf(2)$$ And we see easily that this lead to a contradiction if $n\to +\infty$.

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    $\begingroup$ Very nice solution. $\endgroup$ – mfl Oct 13 '14 at 16:58
  • $\begingroup$ @mfl Why did you retract your answer? Why does not it work? $\endgroup$ – mathematiccian Oct 13 '14 at 18:37
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    $\begingroup$ Because, for example, $f''(x)=0$ not only for $x=0.$ The same happens with higher order derivatives. Thus it does not satisfy the your second requirement. $\endgroup$ – mfl Oct 13 '14 at 18:41
  • $\begingroup$ @Kelenner In Taylor Lagrange formula does the same c make the identity hold for all n? So there exists a c such that the identity holds for all n. It is not the case that for all n there exists a different c? It does not matter anyway for the proof, right? $\endgroup$ – mathematiccian Oct 13 '14 at 18:53
  • $\begingroup$ @mathematician The $c$ are in fact (in general) dependant of $x$ and $n$. But the important fact is that it belong to $(0,x)$ in the last part, so as I have taken $x\in (0,1)$, $f^{(n)}(c)\leq f^{(n)}(1)$ as $f^{(n)}$ is increasing. $\endgroup$ – Kelenner Oct 13 '14 at 19:33

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