Separate zeros of a smooth function This question occurred to me while working on paper. I would like to separate a zero from a smooth function similar to what you can do with polynomials. So here is my full question:

If I am given a smooth function $f: \mathbb{R}^n \to \mathbb{R} $  which vanishes in $0$ and in some further derivatives, i.e.
$$ f(0) =0 \text{ and  }  \frac{\partial^\alpha f}{ \partial x^\alpha } (0) =0,$$
Can I conclude that there is a multiindex $\tilde\alpha \neq \alpha $ such that I can write $f$ in terms of
$$ f = x^\tilde\alpha  h $$ for a smooth function $h : \mathbb{R}^n \to \mathbb{R}$?

Can anyone explain why this is possible or why not?
No you cannot conclude that. A smooth function may vanish to infinite order at a single point, which means that all the derivatives can vanish at once, without the function itself being identically zero. The classic example is $\exp(-1/x^2)$ for positive $x$ and 0 for non-positive $x$.
No you cannot conclude that. A smooth function may vanish to infinite order at a single point, which means that all the derivatives can vanish at once, without the function itself being identically zero. The classic example is $\exp(-1/x^2)$ for positive $x$ and 0 for non-positive $x$.