Non-analytic smooth function

The Wikipedia page (http://en.wikipedia.org/wiki/Non-analytic_smooth_function) proves that $$f(x) = \begin{cases} \exp(-1/x), & \mbox{if }x>0 \\ 0, & \mbox{if }x\le0 \end{cases}$$ is a non-analytic smooth function. But I don't understand why the fact that "the Taylor series of $f$ at the origin converges everywhere [where does "everywhere" mean here?] to $0$" implies the Taylor series converges to $0$ when $x>0$? Can someone explain this in greater detail? I guess there should be a gap in my understanding of power series.

• The Taylor series of that function at 0 is $0+0x+0x^2/2!+0x^3/3!+\cdots$, with all terms zero. Therefore it converges for all $x$, that is, «everywhere». Jan 30 '15 at 19:24
• @MarianoSuárez-Alvarez If a real analytic function is 0 in an open set in $\mathbb{R}$, can I say it's 0 on the whole real line by this argument? Jan 30 '15 at 19:46
• Yes. That is a basic fact of analytic functions. Jan 30 '15 at 20:03

One can form the Taylor series associated to $f(x)$ centered at $0$, $$\sum_{n = 0}^\infty \frac{f^{(n)}(0)}{n!} x^n \tag{1}.$$ We can talk about the convergence or diververgence of the power series expansion in $(1)$ as a function of $x$, and this is possibly different than considering the behavior of the function $f(x)$ itself. In this case, it happens to be that $f^{(n)}(0) = 0$ for all $n \geq 0$, so the power series is identically zero for all $x$. This is what is meant in the article by "the Taylor series... converges everywhere to 0."

This shows that $f$ is not analytic, as $f$ does not agree with its Taylor series centered at $x=0$ in any open neighborhood of $0$. [Namely, $f(x) > 0$ for $x > 0$, whereas the Taylor series is identically zero].

1. The power series $$\sum_{n=0}^{\infty} 0 \, x^n = 0 + 0x + 0x^2 + 0x^3 + \dots$$ clearly converges to zero for any value of $$x$$. That is, if we call this power series $$g(x)$$, then we can say $$g(x) = 0$$ for any $$x \in \mathrm{I\!R}$$ (another way of saying this is that the radius of convergence for this series is infinite; the important thing to notice is that this is true for any value of $$x$$).

2. $$\vphantom{^{\Big|}}$$ The Taylor series of a function $$h(x)$$ about $$x = c$$ is given by $$\sum_{n=0}^{\infty} \tfrac{h^{(n)}(c )}{n!}(x-c)^n = h(c) + \tfrac{h'(c )}{1!}(x-c)+ \tfrac{h''(c )}{2!}(x-c)^2 + \dots \;,$$ i.e. the coefficients are determined by the derivatives of $$f(x)$$. Analytic functions are equal to their Taylor series expansion wherever the series converges (i.e. over the interval of convergence).

3. $$\vphantom{^{\Big|}}$$ The derivatives (of all order) at $$c = 0$$ of the function $$f(x)$$ given in the OP are all zero (the referenced Wiki page provides a proof), i.e.: $$f^{(n)}(0) = 0, \qquad n = 1, 2, 3, \dots$$ So the coefficients of its Taylor series centred at $$c = 0$$ are given by:
$$\frac{f^{(n)}(0)}{n!} = \frac{0}{n!} = 0, \qquad n = 1, 2, 3, \dots$$ That is, we have shown that the Taylor series for $$f(x) = 0 + 0x + 0x^2 + 0x^3 + \dots$$.

That is, the Taylor series expansion for $$f(x)$$ about $$c = 0$$ is clearly equal to the power series $$g(x)$$ discussed above (1.), which we observed has infinite interval of convergence (i.e. converges for all real $$x$$).

However, it is also clear that $$f(x)$$ is not equal to its Taylor series, because clearly: $$f(x) = \begin{cases}0, & x \leq 0 \\ e^{-1/x} & x > 0 \end{cases} \quad \mathbf{{\color{red} \neq }} \quad 0 + 0x + 0x^2 + \dots$$

It sounds to me as though you were conflating the centre about which we make the expansion $$x = c$$ and the value at which we evaluate the function $$x$$ in $$f(x)$$ or $$g(x)$$.

The point of the example is not that $$f(x)$$ doesn't have an accurate series representation of some sort (over some interval) around any point, but that the expansion about $$c = 0$$,$$^{\color{red}{1}}$$ is well-defined for all values of $$x$$, and yet it is not equal to $$f(x)$$. In other words, there do exist well-defined, convergent Taylor expansions of a function $$f(x)$$ that do not converge to $$f(x)$$.

If this feels kind of contrived / unsatisfying / underwhelming, you're not alone :). From what I understand, it is an important result, but hard to appreciate if your experience is limited to analytic functions.

$$^{\color{red}1}$$: It is possible to find an accurate series expansion around other points, although they wouldn't be "Taylor" series (because the powers of $$x$$ wouldn't be positive integers).