MHB Uniform convergence of a complex power series on a compact set

Click For Summary
The discussion focuses on proving the uniform convergence of the complex power series $\sum_{n=0}^{\infty} a_n z^n$ on the compact disc defined by $|z| \leq r|z_0|$, given that the series converges for some $z_0 \neq 0$. It is established that the series converges absolutely for $|z| < |z_0|$, leading to the conclusion that the terms $|a_n z^n|$ can be bounded by a geometric series. The key challenge is demonstrating that for any $\epsilon > 0$, there exists an integer $N$ such that $|a_{n+1} z^{n+1}| < \epsilon$ for all $n > N$. The discussion emphasizes the need to show this condition for uniform convergence across the compact set. Ultimately, the proof hinges on establishing appropriate bounds for the terms of the series.
kalish1
Messages
79
Reaction score
0
I need to prove that the complex power series $\sum\limits_{n=0}^{\infty}a_nz^n$ converges uniformly on the compact disc $|z| \leq r|z_0|,$ assuming that the series converges for some $z_0 \neq 0.$

*I know that the series converges absolutely for every $z,$ such that $|z|<|z_0|.$ Since $\sum\limits_{n=0}^{\infty}a_nz_0^n$ converges, by definition of convergence this means that the tail part of the series can be made arbitrarily small. So $\lim\limits_{n \rightarrow \infty}a_nz_0^n = 0$ and this means that $\exists M \in \mathbb{R}^{+}$ so that $|a_nz_0^n| \leq M$ for $n \geq M.$ So it follows that $$|a_nz^n| \leq M \left|\dfrac{z}{z_0}\right|^n.$$ Thus $$\sum\limits_{n=0}^{\infty}|a_nz^n| \leq \sum\limits_{n=0}^{\infty}M \left|\dfrac{z}{z_0}\right|^n = M\sum\limits_{n=0}^{\infty}r^n,$$ where $r<1$ (because $|z|<|z_0| \implies \left|\dfrac{z}{z_0}\right|<1.$)*

But how do I get started on this uniform convergence?
 
Physics news on Phys.org
How can I show that for all $\epsilon > 0$ there exists $N \in \mathbb{N}$ so that $|a_{n+1}z^{n+1}|<\epsilon$ for all $n>N$?
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K