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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3As in $f(x) = x$? – user795305 Oct 10 '14 at 01:19
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Question is edited – mathematiccian Oct 10 '14 at 01:55
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I think that considering the Taylor series of such a function would show that the desired function has to be identically zero, contrary to your hopes. – user795305 Oct 10 '14 at 01:57
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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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