# 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?

• How would you do this for $p(x,y) = x + y^2?$ – zhw. Aug 21 '16 at 14:40

## 1 Answer

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 $x$.

• Also see "Malgrange Preparation Theorem" for what you can conclude if you assume there is a non-vanishing derivative. – Thompson Aug 21 '16 at 14:44
• You can still write your example as $xg(x),$ where $g$ is smooth. – zhw. Aug 21 '16 at 16:28