Let $g(x)=e^{-1/x^2}$ for $x\neq0$, and $g(0)=0$. I've shown that $g^{(n)}(0)=0$ and thus that it is infinitely differentiable about 0, but now I must show that it cannot be expressed as a power series about 0.
The proof seems easy enough using Taylor series, but we haven't covered these in class, so it seems like a bad idea to try and use them. Is there a simple way without mentioning these?
The answer is still the same. If we had $g(x)=\sum_{n=0}^\infty c_n x^n$ (convergent in a neighbourhood of $0$), then $g^{(n)}(0)=n!c_n$, hence all $c_n=0$, but tat clearly differs from $g$.