TITLE: Are there any $f(x)$ whose $\xi_n$ of the Lagrange's remainder does not converge to $0$?
QUESTION [6 upvotes]: Consider the Maclaurin's series of an analytic function $f$ with Lagrange's remainder.
Define $\xi_n$ as
$$f(1)=\sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}+\frac{f^{(n)}(\xi_n)}{n!}$$
where $0<\xi_n<1$, and if there are several that satisfy this, choose the one closest to $0$ (or the infimum if there are infinitely many).
Is there an example of $f$ such that $\xi_n$ doesn't converge to $0$?
I'm not sure why, but all the ones I tried seemed to go to zero.

Edit:
$\frac{f^{(n)}(\xi_n)}{f^{(n)}(0)}=1+\frac{1}{n+1}\frac{f^{(n+1)}(\xi_{n+1})}{f^{(n)}(0)}$, so, if $\frac{f^{(n+1)}(\xi_{n+1})}{f^{(n)}(0)}=o(n)$ then $f^{(n)}(\xi_n)\to f^{(n)}(0)$. This implies that $\xi_n$ converges to $0$ for some functions such that $e^{ax}, \sin ax$. From the results of numerical methods for other functions, I suppose that the same is true for all analytic functions.

REPLY [2 votes]: Let $f$ be a compactly supported smooth function with compact support $K$ such that $1\in K$, $0\notin K$ and $f(1)>0$ (such function is easy to construct). Choose a sequence $\xi_n$ such that (there exists at least one in virtue of Taylor's theorem)
$$f(1)=\sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}+\frac{f^{(n)}(\xi_n)}{n!}=\frac{f^{(n)}(\xi_n)}{n!},\quad  \forall n\geq 0.$$
Then, $\xi_n$ cannot converge to $0$ because, as $f^{(n)}$ is identically $0$ in a neighborhood (which does not depends on $n$) of $0$, this would lead to $f(1)=0$.
Of course, $f$ cannot be analytic. So it do not fully answer your question but you seemed interested in non-analytic examples.