InfinitelyDifferentiableFunctionThatIsNotAnalytic 2002-06-09 09:40:13 2002-06-09 09:52:23 Example analytic % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\Prob}[2]{\mathbb{P}_{#1}\left\{#2\right\}} \newcommand{\Expect}{\mathbb{E}} \newcommand{\norm}[1]{\left\|#1\right\|} % Some sets \newcommand{\Nats}{\mathbb{N}} \newcommand{\Ints}{\mathbb{Z}} \newcommand{\Reals}{\mathbb{R}} \newcommand{\Complex}{\mathbb{C}} %%%%%% END OF SAVED PREAMBLE %%%%%% \usepackage{amsthm} If $f\in\mathcal{C}^{\infty}$, then we can certainly \emph{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. \newcommand{\e}{e^{-\frac{1}{x^2}}} Let $$ f(x)=\begin{cases} \e & x \ne 0 \\ 0 & x = 0 \end{cases}. $$ 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 $f$. \subsection*{Proof that $f^{(n)}(0)=0$} Let $p(x), q(x)\in \Reals[x]$ be polynomials, and define $$ g(x)=\frac{p(x)}{q(x)} \cdot f(x). $$ Then, for $x\ne 0$, $$ g'(x) = \frac{(p'(x) + p(x)\frac{2}{x^3})q(x) - q'(x)p(x)}{q^2(x)} \cdot\e. $$ Computing (e.g. by applying \PMlinkname{L'H\^opital's rule}{LHpitalsRule}), 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.