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.
Best Answer
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.