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.

  • 4
    $\begingroup$ 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 $0$. $\endgroup$ – Jean-Claude Arbaut Sep 23 '16 at 14:20
  • 4
    $\begingroup$ 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 $1$. $\endgroup$ – Jean-Claude Arbaut Sep 23 '16 at 14:27
  • $\begingroup$ 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? $\endgroup$ – user55912 Sep 23 '16 at 14:32
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.