TITLE: What if $f^{(n)}(a)=0$ for all $n\geq 0$?
QUESTION [7 upvotes]: This morning I was trying to imagine what a function would look like if all it's derivatives were zero at a point $a$ (assuming it is $C^\infty$). My first thought was that it should be identically zero in a neighborhood of $a$, but this is only true if the function is analytic. So my question is this:

What can we say in general about a smooth function (i.e. has derivatives of all orders) whose derivatives are all zero at a single point $a$?

It seems like the function $f$ should approach zero at $a$ faster than any polynomial because of Taylor's theorem. In other words, for every $k\geq 1$, there exists some remainder function $r_k(x)$ with $$\frac{f(x)}{(x-a)^k}=r_k(x)\qquad \left( \text{and}\quad \lim_{x\rightarrow a}\ r_k(x)= 0\right).$$
So can we conclude that $f$ approaches zero faster than any polynomial? If so, how is this made precise? And is this all we can say about these functions? (These questions can all be subordinated to the main question)

REPLY [5 votes]: If all the derivatives of $f$ at $0$ vanish, then $\lim_{x \to 0}\  f(x)/x^k=0$ for all $k$, by l'Hospital's theorem used $k$ times.
The converse is not true: $e^{-1/x^2} \sin e^{1/x^2}$ goes to zero faster than any polynomial, but its derivative is not defined continuous at $0$, so the second derivative at $0$ doesn't exist.
If $f$ is infinitely differentiable at $0$, and goes to $0$ faster than any polynomial, then $f$ has all its derivatives at $0$ vanish; this follows from Taylor's theorem. I'll give a bit more detail: Suppose for the sake of contradiction that $f$ is infinitely differentiable and $f^{(k)}(0) = c \neq 0$. Then Taylor's theorem shows that $f(x) = c x^k/k! +  r(x)$ where $\lim_{x \to 0} r(x)/x^k=0$. So $\lim_{x \to 0} f(x)/x^k = c/k!$, not $0$, a contradiction.