3.8 KiB
Lecture 8
Review
Sequences of Functions
Let f_n: G \to \mathbb{C} be a sequence of functions.
Convergence Pointwise
Definition:
Let \zeta\in G, \forall \epsilon > 0, \exists N \in \mathbb{N} such that \forall n \geq N, |f_n(\zeta) - f(\zeta)| < \epsilon.
Convergence Uniformly
Definition:
\forall \epsilon > 0, \forall \zeta\in G, \exists N \in \mathbb{N} such that \forall n \geq N, |f_n(\zeta) - f(\zeta)| < \epsilon.
Convergence Locally Uniformly
Definition:
\forall \epsilon > 0, \forall \zeta\in G, \exists N \in \mathbb{N} such that \forall n \geq N, |f_n(\zeta) - f(\zeta)| < \epsilon.
Convergence Uniformly on Compact Sets
Definition: \forall C\subset G that is compact, \forall \epsilon > 0, \exists N \in \mathbb{N} \text{ s.t. } \forall n \geq N, \forall \zeta\in C, |f_n(\zeta) - f(\zeta)| < \epsilon
Power Series
Definition:
\sum_{n=0}^{\infty} c_n (\zeta - \zeta_0)^n
\zeta_0 is the center of the power series.
Theorem of Power Seriess
If a power series converges at \zeta_1, then it converges absolutely at every point of \overline{B(0,r)} that is strictly inside the disk of convergence.
Continue on Power Series
Limits of Power Series
Theorem 5.12
Cauchy-Hadamard Theorem:
The radius of convergence of the power series is given by \sum_{n=0}^{\infty} a_n (\zeta - \zeta_0)^n is given by
\frac{1}{R} = \limsup_{n\to\infty} |a_n|^{1/n}
Proof:
Suppose (b_n)^{\infty}_{n=0} is a sequence of real numbers such that \lim_{n\to\infty} b_n may nor may not exists by (-1)^n(1-\frac{1}{n}).
The limit superior of (b_n) is defined as
s_n = \sup_{k\geq n} b_k
s_n is a decreasing sequence, by completeness of \mathbb{R}, every bounded sequence has a limit in \mathbb{R}.
So s_n converges to some limit s\in\mathbb{R}.
Without loss of generality, this also holds for infininum of s_n.
Forward direction:
We want to show that the radius of convergence of \sum_{n=0}^{\infty} a_n (\zeta - \zeta_0)^n is greater than or equal to \frac{1}{\limsup_{n\to\infty} |a_n|^{1/n}}.
Since \sum_{n=0}^{\infty} 1\zeta^n=\frac{1}{1-\zeta} for |\zeta|<1. Assume \limsup_{n\to\infty} |a_n|^{1/n} is finite, then \sum_{n=0}^{\infty} a_n (\zeta - \zeta_0)^n converges absolutely at \zeta_0.
Let \rho>\limsup_{n\to\infty} |a_n|^{1/n}, then \exists N \in \mathbb{N} such that \forall n \geq N, |a_n|^{1/n}\leq \rho.
So \frac{1}{R}=\limsup_{n\to\infty} |a_n|^{1/n}<\rho
So R>\frac{1}{\rho}
/TRACK LOST/
Backward direction:
Suppose |\zeta|>R, then \exists number |\zeta| such that |\zeta|>\frac{1}{\rho}>R.
So \rho<\limsup_{n\to\infty} |a_n|^{1/n}
This means that \exists infinitely many $n_j$s such that |a_{n_j}|^{1/n_j}>\rho
So |a_{n_j}\zeta^{n_j}|>\rho^{n_j}|\zeta|^{n_j}
Series \sum_{n=1}^{\infty} a_n\zeta^n diverges, each individual term is not going to 0.
So \sum_{n=0}^{\infty} a_n (\zeta - \zeta_0)^n does not converge at \zeta
EOP
What if |\zeta-\zeta_0|=R?
For \sum_{n=0}^{\infty} \zeta^n, the radius of convergence is 1.
It diverges eventually on the circle of convergence.
For \sum_{n=0}^{\infty} \frac{1}{(n+1)^2}\zeta^n, the radius of convergence is 1.
This converges everywhere on the circle of convergence.
For \sum_{n=0}^{\infty} \frac{1}{n+1}\zeta^n, the radius of convergence is 1.
This diverges at \zeta=1 (harmonic series) and converges at \zeta=-1 (alternating harmonic series).
Theorem 5.15
Suppose \sum_{n=0}^{\infty} a_n (\zeta - \zeta_0)^n has a positive radius of convergence R. Define f(\zeta)=\sum_{n=0}^{\infty} a_n (\zeta - \zeta_0)^n, then f is holomorphic on B(0,R) and f'(\zeta)=\sum_{n=1}^{\infty} n a_n (\zeta - \zeta_0)^{n-1}=\sum_{k=0}^{\infty} (k+1)a_{k+1} (\zeta - \zeta_0)^k.
Proof:
/TRACK LOST/