Maclaurin expansion of zero I have heard that there is a function $f(x)$ whose Maclaurin expansion is zero but f(x) is not identical to zero.
In other words, there exists a function $f(x)$ that satisfies $f^{(n)}(0)$ is zero for every whole number $n$ and $f(x)\neq0$.
Could you show me one of such functions? (I've heard it exists but I've never seen one)
 A: If we take $f(x) = \frac{P(x)}{Q(x)}e^{-1/x^2}$ for $x \neq 0$, then we have (for $x \neq 0$) $$f'(x) = \frac{P'(x)Q(x) - P(x)Q'(x)}{Q(x)^2}e^{-1/x^2} + \frac{2P(x)}{x^3Q(x)}e^{-1/x^2}$$
which assures us that $f'(x)$ will be of the same form (namely, a rational function times $e^{-1/x^2}$). Meanwhile, we know that $f'(0) = \lim_{x \to 0} \frac{P(x)}{xQ(x)}e^{-1/x^2},$ but given that the exponential asymptotically dominates polynomials and that $e^{-1/x^2} \to 0$ as $x \to 0$, we know that regardless of $xQ(x) \to 0$ as $x \to 0$, we must have $f'(0) = 0.$
It follows by induction that $f(x) = e^{-1/x^2}$ for $x \neq 0$ and $f(0) = 0$ satisfies $f^{(n)}(0) = 0$. Thus $f(x)$ can't be computed by its Taylor series. 
A: Yes you appear to be talking about bump functions.
Contrast the complex case, where differentiable  (holomorphic) implies (complex) analytic.   The bump function,  on the other hand, doesn't equal its Taylor series at O (MacLaurin series ), though it's smooth.   If it did it would be zero. ..
