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Does this definition contain redundancy: "A function $f$ is said to be analytic at $x_0$ if its Taylor series about $x_0$ exists and converges to $f(x)$ for all $x$ in some interval containing $x_0$". Is anything missing in the following definition: "A function $f$ is said to be analytic at $x_0$ if it has derivatives of all orders at $x_0$".

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The function $\begin{cases}e^{-\frac{1}{x}}&x>0\\0 & x\leq0\end{cases}$ has derivatives of all order at $x_0=0$, but this function is not analytic. If it were analytic, then a power series would converge to identically zero, since $f$ is identically zero for $x\leq0$.

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You can think about the definition of an analytic function (at $x_0$) as consisting of three parts:

  1. The derivatives of $f$ of all orders at $x_0$ exist. This is equivalent to saying "The Taylor series around $x_0$ exists" (because to define the taylor series around $x_0$, you need to know the derivatives of all orders at $x = x_0$).
  2. The Taylor series converges. This means that the Taylor series has a positive radius of convergence. Even if the deriatives of all orders at $x = x_0$ exists, this doesn't mean that the Taylor series will have a positive radius of convergence (see Borel's lemma).
  3. The Taylor series converges to $f$. Even if the Taylor series has a positive radius of convergence, it might converge to a function different than $f$ as it happens in the examples provided in the other answers.

Your attempt of simplification only captures the first part.

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    $\begingroup$ This reminds me of the standard elementary calculus definition of continuity at a point. The function has to be defined at the point, the limit of the function has to exist at the point, and these two values have to be equal. $\endgroup$ – Dave L. Renfro Dec 8 '15 at 20:51
  • $\begingroup$ This is probably splitting hairs, but I feel like saying "the Taylor series exists" is saying it is analytic, since in Taylor's theorem we need the remainder term to go to zero for the series to converge? I guess this all depends on how you define the "Taylor series". $\endgroup$ – charlestoncrabb Dec 8 '15 at 21:19
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    $\begingroup$ It might depend on how you define "the Taylor series converges", not how you define the Taylor power series (as a formal series). Taylor's theorem helps us to relate the Taylor polynomials to the original function but it doesn't force the remainder to go to zero for the series to converge. The Taylor power series can converge without the remainder (as in, the difference between the Taylor polynomials and the function, not the tail of the Taylor series) going to zero. This is precisely what happens in the examples below. $\endgroup$ – levap Dec 8 '15 at 21:32
  • $\begingroup$ The Taylor series converges to the zero function and the remainder which we expect to go to zero converges instead to the original function. $\endgroup$ – levap Dec 8 '15 at 21:32
  • $\begingroup$ @charlestoncrabb, the Taylor series as a formal power series is a useful tool to handle all derivatives at once, even if does not converge or not to the function. $\endgroup$ – Carsten S Dec 9 '15 at 0:07
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$x\mapsto f(x)=\exp(-1/x^2)$ extended to $f(0)=0$ has all its derivatives equal to 0 at 0 so it cannot be equal to its own Taylor series in any interval around 0.

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