0
$\begingroup$

We know that a function f(z) is analytic iff it can be given by the locally convergent power series. If it is analytic everywhere it must be convergent for all z. What about the analyticity of the function outside the circle of convergence and on the circle of convergence? How to evaluate the singularities from such a power series expansion of the function?

$\endgroup$
  • $\begingroup$ what? if it is convergent for all $z$, then the radius of convergence is infinite, and so there's really no circle here... $\endgroup$ – mathworker21 May 15 at 5:38
  • $\begingroup$ The question I mean is specified when the function is not analytic everywhere. $\endgroup$ – MB17 May 15 at 5:46
  • 1
    $\begingroup$ The definition of "$f$ is analytic at $z$" is that $f$ is given by a convergent power series in some neighborhood of $z$, equivalently, that it has a power series centered at $z$ with a positive radius of convergence. However, a finite radius of convergence at one value of $z$ doesn't preclude the function from being analytic outside that circle; in other words, "the circle of convergence" isn't well defined. Consider for example $f(z)=1/(1-z)$, which has a power series $\sum_{n=0}^\infty z^n$ in the open unit disk, but still is analytic everywhere except $z=1$. $\endgroup$ – Greg Martin May 15 at 6:16
0
$\begingroup$

Let $f : U \to \mathbb C$ be analytic on an open $U \subsetneqq \mathbb C$. For $z_0 \in U$ let $r = \inf \{ \lvert v - z_0 \rvert \mid v \in \mathbb C \setminus U \}$. Then $U_r(z_0) = \{ z \in \mathbb C \mid \lvert z - z_0 \rvert < r \}$ is the biggest open disk centered at $z_0$ which is contained in $U$. Note that the boundary $S_r(z_0) = \{ z \in \mathbb C \mid \lvert z - z_0 \rvert = r \}$ contains at least one point not belonging to $U$.

It is well-known that there exists a power series $\sum_{n=0}^\infty a_n(z-z_0)^n$ with radius of convergence $R \ge r$ such that $f(z) = \sum_{n=0}^\infty a_n(z-z_0)^n$ for $z \in U_r(z_0)$.

It is possible that $R > r$ (as a simple example let $f$ be the restriction of an analytic $g : \mathbb C \to \mathbb C$ to an open $U \subsetneqq \mathbb C$). In this case, what can be said about $f(z)$ and $\sum_{n=0}^\infty a_n(z-z_0)^n$ for $z \in (U_R(z_0) \cap U) \setminus U_r(z_0)$? Note that $U_R(z_0) \cap U$ is the set of points where both $f(z)$ and $\sum_{n=0}^\infty a_n(z-z_0)^n$ are defined.

It is very well possible that $f(z) \ne \sum_{n=0}^\infty a_n(z-z_0)^n$. As an example consider the principal branch of the logarithm $\ln: \mathbb C_{sliced} = \mathbb C \setminus \{ x \in \mathbb R \mid x < 0 \} \to \mathbb C$. For $z_0 = -4 + 3i$ we have $r= 4$ and $\ln(z) = \sum_{n=0}^\infty a_n (z-z_0)^n$ for $z \in U_4(z_0)$. However, the radius of convergence of the power series is $R = 5$, but for $z \in U_5(z_0)$ with $\text{Im}(z) < 0$ we have $\ln(z) \ne \sum_{n=0}^\infty a_n (z-z_0)^n$.

However, if $V$ is any connected component of $U \cap U_R(z_0)$ such that$V \cap U_r(z_0) \ne \emptyset$, then $f(z) = \sum_{n=0}^\infty a_n (z-z_0)^n$ for $z \in V$. This is due to the identity theorem for analytic functions.

$\endgroup$
-1
$\begingroup$

Your question is unclear but I have a few advices :

List the power series of $\frac{1}{z}$ at every point $z \ne 0$ and their radius of convergence and how you can find one from the other. Do the same with its primitive $\log z$.

The main theorem is the Cauchy integral formula : if $f$ is analytic on $|z| < r$ (note a priori $f$ is not given by a single power series on that disk) then for $|z|< R< r$, $f(z) = \frac{1}{2i\pi}\int_{|s|=R} \frac{f(s)}{s-z}ds$, expanding $\frac{1}{1-z/s}$ in geometric series shows $f$ is indeed given by a single power series.

From there we obtain the second main theorem : (the analytic continuation of) a power series always has a singularity of the boundary of its disk of convergence.

The 3rd theorem is that the Cauchy integral formula holds for holomorphic functions : holomorphic functions are analytic (from which we obtain things like the removable singularity theorem).

$\endgroup$

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.