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[parent] infinitely-differentiable function that is not analytic (Example)

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

$\displaystyle f(x)=\begin{cases} e^{-\frac{1}{x^2}}& x \ne 0 \ 0 & x = 0 \end{cases}. $
Then $ f\in \mathcal{C}^{\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 \mathbb{R}[x]$ be polynomials, and define

$\displaystyle g(x)=\frac{p(x)}{q(x)} \cdot f(x). $
Then, for $ x\ne 0$,
$\displaystyle g'(x) = \frac{(p'(x) + p(x)\frac{2}{x^3})q(x) - q'(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'(0)=\lim_{x\to 0}g'(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.



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Cross-references: polynomials, power series, necessary, radius of convergence, converge, Taylor series
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This is version 2 of infinitely-differentiable function that is not analytic, born on 2002-06-09, modified 2002-06-09.
Object id is 3081, canonical name is InfinitelyDifferentiableFunctionThatIsNotAnalytic.
Accessed 5824 times total.

Classification:
AMS MSC30B10 (Functions of a complex variable :: Series expansions :: Power series )
 26A99 (Real functions :: Functions of one variable :: Miscellaneous)

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