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.


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 $n\to +\infty$.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.