All Questions
7
questions
3
votes
1
answer
132
views
Can this given $f: S^1\to \mathbb C$ be extended to a continuous $F: \overline{\mathbb D}\to \mathbb C, F$ is holomorphic on $\mathbb D$?
Suppose that $f: \mathbb S^1\to \mathbb C$ is continuous such that $f(z)=f(\bar z)$ for all $z\in \mathbb S^1$. Can it be extended to a continuous $F: \overline{\mathbb D}\to \mathbb C$ such that $ F$ ...
- 11.5k
0
votes
0
answers
86
views
Continuity on boundary of convergence power series
I’m stuck trying to prove the following: Given $f(z) = \log(2+z^2)$, I consider its power series representation around $z=0$, which is $i(z) = \log(2)+\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n2^n}z^{2n}$...
- 1
1
vote
0
answers
143
views
Continuity at the boundary of a convergent power series
Say $f(z)=\sum a_{n}z^{n}$ is a power series with convergence radius $0<R<\infty$. Suppose we know that the series convergence at $z_{0}$ where $z_{0}$ is a point at the boundary of the ...
- 1,501
0
votes
1
answer
40
views
Showing that $f(z) = \sum_{n=1}^{\infty} (n+1)^2 z^n$ is continuous in the open unit disk $|z| <1$
So I figured since its a power series that maybe finding the radius of convergence here would be useful. So as $f(z) = \sum_{n=1}^{\infty} (n+1)^2 z^n$ we can find the radius of convergence as$$R= \...
- 904
0
votes
1
answer
83
views
Power series of analytic function
Suppose $f(z)=\sum\limits_{n=0}^{\infty}a_n z^n$ for $|z|<r$, where $r>0$.
Also $f(z)$ is continuous on $|z|\leq r$.
My question is whether the function $f(z)$ is analytic in some "big" disk $|z|...
- 8,455
5
votes
2
answers
1k
views
Holomorphic function on unit disc has no continuous extension to boundary
The following is a qual study question:
Let $$f(z) = \sum_{n=1}^\infty \sqrt{n}z^n$$
Having proven that the radius of convergence is 1, I'm asked to show that this function cannot be extended to a ...
- 287
2
votes
1
answer
241
views
Show that $f(z):=\sum a_n (z-z_0)^n$ is continuous whenever $z$ is in disk of convergence.
Consider a power series $\sum a_n(z-z_0)^n$, and assume it has radius of convergence $r$. Then we know that $\forall z\in(z_0 -r,z_0 +r)$, this power series converges absolutely by root test. Thus we ...
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