infinitely-differentiable function that is not analytic

If $f\in {𝒞}^{\mathrm{\infty }}$, then we can certainly write a Taylor series for $f$. However, analyticity requires that this Taylor series actually converge (at least across some radius of convergence) to $f$. It is not necessary that the power series for $f$ converge to $f$, as the following example shows.

Let

Then $f\in {𝒞}^{\mathrm{\infty }}$, and for any $n\ge 0$, $f^{(n)}(0)=0$ (see below). So the Taylor series for $f$ around 0 is 0; since $f(x)>0$ for all $x\ne 0$, clearly it does not converge to $f$.

Proof that $f^{(n)}(0)=0$

Let $p(x),q(x)\in ℝ[x]$ be polynomials, and define

$$g(x)=\frac{p(x)}{q(x)}\cdot f(x).$$

Then, for $x\ne 0$,

$$g^{\prime }(x)=\frac{(p^{\prime }(x)+p(x)\frac{2}{x^3})q(x)-q^{\prime }(x)p(x)}{q^2(x)}\cdot e^{-\frac{1}{x^2}}.$$

Computing (e.g. by applying L’Hôpital’s rule (http://planetmath.org/LHpitalsRule)), we see that $g^{\prime }(0)={lim}_{x\to 0}{g}^{\prime }(x)=0$.

Define $p_0(x)=q_0(x)=1$. Applying the above inductively, we see that we may write $f^{(n)}(x)=\frac{p_n(x)}{q_n(x)}f(x)$. So $f^{(n)}(0)=0$, as required.