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# infinitely-differentiable function that is not analytic

If $f\in\mathcal{C}^{{\infty}}$, 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

$f(x)=\begin{cases}e^{{-\frac{1}{x^{2}}}}&x\neq 0\\ 0&x=0\end{cases}.$ |

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

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

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

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

Then, for $x\neq 0$,

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

Computing (e.g. by applying L’Hôpital’s rule), 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.

## Mathematics Subject Classification

30B10*no label found*26A99

*no label found*

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