Is there a smooth function $f$ that for all $n\in\mathbb{Z}_+$, $f^{(n)}(n)=0$ i.e. $n$th derivative at the point $n$ is zero and $f^{(n)}(x)\ne 0$ for all $x\in\mathbb R\setminus \{n\}$? If there is that kind of function, can anyone give an example of such a function?
$\begingroup$
$\endgroup$
$\begingroup$
$\endgroup$
Let $$\varphi (t) =\begin{cases} e^{\frac{1}{\left(x-\frac{1}{2}\right)^2 -\frac{1}{4}}} \mbox{ if } 0<x<1 \\ 0 \hspace{1.2cm} \mbox{ if } x\in \mathbb{R}\setminus (0,1) \end{cases} $$ And take $$f(x) =\sum_{n\in\mathbb{Z}} \varphi (n+x) .$$
answered May 29 '15 at 19:34
-
$\begingroup$ Doesn't this mean that $f(n)=0$ for all $n\in\mathbb{Z}$? I mean that the function and every derivative of it should have only one root. $\endgroup$ – guest May 30 '15 at 19:35
