# Definition for real analytic 'function'

I'm studying analytic functions out of Rudin PMA and the wikipedia article, but I'm not clear on analytic functions. Rudin says a function $f: \mathbb{R} \rightarrow \mathbb{R}$ is analytic at $x = a$ if there exists $\{c_n\} \subset \mathbb{R}$ and $R > 0$ such that $f(x) = \sum c_{n}(x - a)^n$ for all $|x - a| < R$.

Now, from Rudin I can see that a function $f$ is real analytic on an open set $D \subset \mathbb{R}$ if it has a power series expansion for each $a \in D$. But what do we talk about if we say a function $f$ is ''real analytic''. Like the definition for continuity of a function, I assume this means that a power series expansion may be found for $f$ at every point in its domain. But Rudin doesn't say this explicitly, so I look at Wikipedia.

''The reciprocal of an analytic function that is nowhere zero is analytic''. For the moment, I believe this statement to be false. Consider $f(x) = \frac{1}{1-x}$, which is the reciprocal of the analytic function $1-x$. Also $1-x$ is nowhere zero if we don't allow $x = 1$ in the domain. From the basic geometric series formula, we have $f(x) = \sum_{n=0}^{\infty}x^{n}$ if and only if $|x| < 1$. We can't express $f(x)$ as a power series if $|x| \geq 1$. In particular, $2$ belongs tot he domain of $f$ but there is no power series in a neighborhood of $2$. So $f$ is not real analytic at every $x$ in the domain of $f$, meaning $f$ is not real analytic.

I appreciate all feedback. Thanks.

• Yes, you can develop $\dfrac{1}{1-x}$ in series around any $x_0$ with $|x_0|>1$. It is said nowhere that it has to be the same series everywhere. The development $\dfrac{1}{1-x}=\sum_{n\geq0}x^n$ is around $0$. – Jean-Claude Arbaut Sep 23 '16 at 14:20
• For instance, as you choose $2$, you have, for $x$ in a neighborhood of $2$: $$\dfrac{1}{1-x}=\dfrac{1}{-1-(x-2)}=\dfrac{-1}{1+(x-2)}=-\sum_{n=0}^{\infty}(-1)^n(x-2)^n$$ Its radius of convergence is $1$. – Jean-Claude Arbaut Sep 23 '16 at 14:27
• Okay, well how do we develop a series for some $|x_0| > 1$? Perhaps we look at $f(x) = \frac{1}{1-x} = \frac{1}{(1 - x_0) - (x - x_0)} = \frac{1}{1 - x_0} \frac{1}{1 - \frac{x - x_0}{1 - x_0}}$ and do some geometric series argument for the right-hand side? – user55912 Sep 23 '16 at 14:32
• Exactly. An interesting point: the radius of this series is $|x_0-1|$, or exactly the distance from $x_0$ to $1$, the pole closest (actually unique here) to $x_0$. See this. – Jean-Claude Arbaut Sep 23 '16 at 14:35

Basically for a (real) function $f$ to be analytic, it has a power series on some neighborhood. So to quote what you have your power series is $\sum c_n(x-a)^n$ and neighborhood $x \in (-a + R, a + R).$ You might think, well surely this ought to be true if the function is smooth right? No, look at this classical example

$$f(x) = \left\{\begin{matrix} e^{-1/x} & x > 0\\ 0& x \leq 0 \end{matrix}\right.$$

Now $\forall n \geq 0$, $f^n(0) = 0$. Hence this is not real analytic as it cannot be equal to its own Taylor series.