Smooth non-analytic function

Counterexample

Let $f(x) = e^{-1/x}$ for $x > 0$ and $f(x) = 0$ for $x \le 0$. Then $f$ is smooth on $\mathbb{R}$, and $f^{(n)}(0) = 0$ for all $n$. So its Taylor series at $0$ is the zero series. But $f(x) > 0$ for $x > 0$, so $f$ is not analytic at $0$.

Remarks

For $x > 0$, each derivative has the form $p_n(1/x) e^{-1/x}$ for a polynomial $p_n$. It tends to $0$ as $x \to 0^+$, because $e^{-1/x}$ decays faster than any power of $x$. Such functions give smooth bump functions and smooth partitions of unity. Mathlib defines this function as expNegInvGlue.

Explanations

L1 human

This function is so flat at $0$ that all its derivatives vanish there. So its Taylor series sees only the zero function, but the function is not zero.