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 (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.

Title infinitely-differentiable function that is not analytic InfinitelydifferentiableFunctionThatIsNotAnalytic 2013-03-22 12:46:15 2013-03-22 12:46:15 ariels (338) ariels (338) 5 ariels (338) Example msc 30B10 msc 26A99