126 lines
3.8 KiB
Markdown
126 lines
3.8 KiB
Markdown
# Math416 Lecture 8
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## Review
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### Sequences of Functions
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Let $f_n: G \to \mathbb{C}$ be a sequence of functions.
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#### Convergence Pointwise
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Definition:
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Let $\zeta\in G$, $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $|f_n(\zeta) - f(\zeta)| < \epsilon$.
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#### Convergence Uniformly
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Definition:
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$\forall \epsilon > 0$, $\forall \zeta\in G$, $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $|f_n(\zeta) - f(\zeta)| < \epsilon$.
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#### Convergence Locally Uniformly
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Definition:
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$\forall \epsilon > 0$, $\forall \zeta\in G$, $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $|f_n(\zeta) - f(\zeta)| < \epsilon$.
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#### Convergence Uniformly on Compact Sets
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Definition: $\forall C\subset G$ that is compact, $\forall \epsilon > 0, \exists N \in \mathbb{N} \text{ s.t. } \forall n \geq N, \forall \zeta\in C, |f_n(\zeta) - f(\zeta)| < \epsilon$
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#### Power Series
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Definition:
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$$
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\sum_{n=0}^{\infty} c_n (\zeta - \zeta_0)^n
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$$
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$\zeta_0$ is the center of the power series.
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#### Theorem of Power Seriess
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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.
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## Continue on Power Series
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### Limits of Power Series
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#### Theorem 5.12
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Cauchy-Hadamard Theorem:
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The radius of convergence of the power series is given by $\sum_{n=0}^{\infty} a_n (\zeta - \zeta_0)^n$ is given by
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$$
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\frac{1}{R} = \limsup_{n\to\infty} |a_n|^{1/n}
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$$
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Proof:
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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 $(-1)^n(1-\frac{1}{n})$.
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The limit superior of $(b_n)$ is defined as
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$$
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s_n = \sup_{k\geq n} b_k
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$$
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$s_n$ is a decreasing sequence, by completeness of $\mathbb{R}$, every bounded sequence has a limit in $\mathbb{R}$.
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So $s_n$ converges to some limit $s\in\mathbb{R}$.
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Without loss of generality, this also holds for infininum of $s_n$.
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Forward direction:
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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 $\frac{1}{\limsup_{n\to\infty} |a_n|^{1/n}}$.
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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 $\zeta_0$.
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Let $\rho>\limsup_{n\to\infty} |a_n|^{1/n}$, then $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $|a_n|^{1/n}\leq \rho$.
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So $\frac{1}{R}=\limsup_{n\to\infty} |a_n|^{1/n}<\rho$
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So $R>\frac{1}{\rho}$
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/*TRACK LOST*/
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Backward direction:
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Suppose $|\zeta|>R$, then $\exists$ number $|\zeta|$ such that $|\zeta|>\frac{1}{\rho}>R$.
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So $\rho<\limsup_{n\to\infty} |a_n|^{1/n}$
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This means that $\exists$ infinitely many $n_j$s such that $|a_{n_j}|^{1/n_j}>\rho$
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So $|a_{n_j}\zeta^{n_j}|>\rho^{n_j}|\zeta|^{n_j}$
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Series $\sum_{n=1}^{\infty} a_n\zeta^n$ diverges, each individual term is not going to $0$.
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So $\sum_{n=0}^{\infty} a_n (\zeta - \zeta_0)^n$ does not converge at $\zeta$
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EOP
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_What if $|\zeta-\zeta_0|=R$?_
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For $\sum_{n=0}^{\infty} \zeta^n$, the radius of convergence is $1$.
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It diverges eventually on the circle of convergence.
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For $\sum_{n=0}^{\infty} \frac{1}{(n+1)^2}\zeta^n$, the radius of convergence is $1$.
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This converges everywhere on the circle of convergence.
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For $\sum_{n=0}^{\infty} \frac{1}{n+1}\zeta^n$, the radius of convergence is $1$.
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This diverges at $\zeta=1$ (harmonic series) and converges at $\zeta=-1$ (alternating harmonic series).
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#### Theorem 5.15
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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 $f'(\zeta)=\sum_{n=1}^{\infty} n a_n (\zeta - \zeta_0)^{n-1}=\sum_{k=0}^{\infty} (k+1)a_{k+1} (\zeta - \zeta_0)^k$.
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Proof:
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/*TRACK LOST*/
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