Non-analytic smooth function
The Wikipedia page () 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.
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$\begingroup$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].
$\endgroup$ $\begingroup$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 $x$).
$\vphantom{^{\Big|}}$ 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).
$\vphantom{^{\Big|}}$ 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).
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