I have a problem understanding the idea behind Analytic functions. (Please correct me on my terminologies while I state my problem).

An analytic function, is a function that has a power series that converges relative to some point $a$. I know that some functions like $f(x)=x$ are analytic no matter which point $x$ is chosen in the domain. The function $f(x)$ is always equal to its Taylor series expansion.

For certain functions, I know that the $x$ you pick must be very specific, relative to a point $a$ you're taking within a 'radius of convergence'. Otherwise the series expansion diverges.

1) When are functions considered 'analytic'? My problem lies in the fact that 'analytic' is referring to the entire function, and yet by definition it relies on the $x$ you pick, and the $a$ you pick. So for the case when there is a radius of convergence, does that mean that a function $f(x)$ is analytic if and only if $x$ is within the radius of convergence?

2) There is a third case when dealing with the convergence of the power series. When $x=a$, the power series trivially converges to 0. Yet, does that mean since it does converge, the particular $f(x)$ taken must be analytic at $x$?

  • $\begingroup$ Are you restricting attention to the complex case or are you also interested in the real case? The answers here are cleaner in the complex case. $\endgroup$ – Ian Sep 1 '15 at 12:54
  • $\begingroup$ Whichever case makes your explanation clearer $\endgroup$ – Malcolm Sep 1 '15 at 12:55

A function $f$ defined on an open set $U$ in $\mathbb{R}$ (resp. $\mathbb{C}$) is called real (resp. complex) analytic if for every $x_0 \in U$ there exists a neighborhood $V$ of $x_0$ such that for any $x \in V$, $f(x)$ can be written as a convergent power series in $x-x_0$. Thus, the dependence on the point of expansion is suppressed through the introduction of this additional neighborhood $V$, which in general depends on $x_0$ (in particular, in general $V$ cannot be taken to be all of $U$).

When we say that a function is analytic at a single point $x$ we really mean that it is analytic on some neighborhood of $x$. Forcing the presence of this open set at all times avoids all the issues you have mentioned.

  • $\begingroup$ This makes a lot more sense! So for example, $f(x)=\frac {1} {x} $ is analytic if it is defined on the domain $\mathbb{R} / {0}$ but not analytical if defined on the domain $\mathbb{R}$? $\endgroup$ – Malcolm Sep 1 '15 at 13:46
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    $\begingroup$ That's kind of a bad example, because that's not even defined at zero. The classic example here is a bump function, which is a smooth function with compact support. Such a function cannot be analytic on any neighborhood containing a boundary point of its support. $\endgroup$ – Ian Sep 1 '15 at 13:54

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