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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3$\begingroup$ As in $f(x) = x$? $\endgroup$– user795305Oct 10, 2014 at 1:19
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$\begingroup$ Question is edited $\endgroup$– mathematiccianOct 10, 2014 at 1:55
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$\begingroup$ 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. $\endgroup$– user795305Oct 10, 2014 at 1:57
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1 Answer
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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.
