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Is there an infinitely differentiable function f : R → R with f(x) = 0 iff x = 0 for which it is reasonable to say the graph of f intersects the x-axis at the origin with infinite multiplicity. So y=x, y=x^2 does not count.

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$f(x) = \begin{cases} e^{-1/x^2}& x \neq 0 \cr 0 & x = 0\end{cases}$

looks like it does the trick. It's a standard proof by induction that it's smooth and that its Maclaurin series is identically zero.

kahen
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