151 lines
4.4 KiB
Markdown
151 lines
4.4 KiB
Markdown
# Lecture 7
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## Review
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### Exponential function
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$$
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e^z=e^{x+iy}=e^x(\cos y+i\sin y)
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$$
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### Logarithm
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#### Definition 4.9 Logarithm
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A logarithm of $a$ is any $b$ such that $e^b=a$.
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### Branch of Logarithm
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A branch of logarithm is a continuous function $f$ on a domain $D$ such that $e^{f(z)}=\exp(f(z))=z$ for all $z\in D$.
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## Continue on Chapter 4 Elementary functions
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### Logarithm
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#### Theorem 4.11
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$\log(\zeta)$ is holomorphic on $\mathbb{C}\setminus\{0\}$.
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Proof:
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We proved that $\frac{\partial}{\partial\overline{z}}e^{\zeta}=0$ on $\mathbb{C}\setminus\{0\}$.
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Then $\frac{d}{dz}e^{\zeta}=\frac{\partial}{\partial x}e^{\zeta}=0$ if we know that $e^{\zeta}$ is holomorphic.
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Since $\frac{d}{dz}e^{\zeta}=e^{\zeta}$, we know that $e^{\zeta}$ is conformal, so any branch of logarithm is also conformal.
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Since $\exp(\log(\zeta))=\zeta$, we know that $\log(\zeta)$ is the inverse of $\exp(\zeta)$, so $\frac{d}{dz}\log(\zeta)=\frac{1}{e^{\log(\zeta)}}=\frac{1}{\zeta}$.
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EOP
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We call $\frac{f'}{f}$ the logarithmic derivative of $f$.
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## Chapter 5. Power series
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If $\sum_{n=0}^{\infty}c_n$ converges, then $\lim_{n\to\infty}c_n=0$ exists.
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### Geometric series
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$$
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\sum_{n=0}^{N}c^n=\frac{1-c^{N+1}}{1-c}
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$$
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If $|c|<1$, then $\lim_{N\to\infty}\sum_{n=0}^{N}c^n=\frac{1}{1-c}$.
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otherwise, the series diverges.
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Proof:
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The geometric series converges if $\frac{c^{N+1}}{1-c}$ converges.
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$$
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(1-c)(1+c+c^2+\cdots+c^N)=1-c^{N+1}
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$$
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If $|c|<1$, then $\lim_{N\to\infty}c^{N+1}=0$, so $\lim_{N\to\infty}(1-c)(1+c+c^2+\cdots+c^N)=1$.
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If $|c|\geq 1$, then $c^{N+1}$ does not converge to 0, so the series diverges.
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EOP
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### Convergence
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#### Definition 5.4
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$$
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\sum_{n=0}^{\infty}c_n
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$$
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converges absolutely if $\sum_{n=0}^{\infty}|c_n|$ converges.
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Note: _Some other properties of converging series covered in Math4111, bad, very bad._
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#### Definition 5.6 Convergence of sequence of functions
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A sequence of functions $f_n$ **converges pointwise** to $f$ on a set $G$ if for every $z\in G$, $\forall\epsilon>0$, $\exists N$ such that for all $n\geq N$, $|f_n(z)-f(z)|<\epsilon$.
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(choose $N$ based on $z$)
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A sequence of functions $f_n$ **converges uniformly** to $f$ on a set $G$ if for every $\epsilon>0$, there exists a positive integer $N$ such that for all $n\geq N$ and all $z\in G$, $|f_n(z)-f(z)|<\epsilon$.
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(choose $N$ based on $\epsilon$)
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A sequence of functions $f_n$ **converges locally uniformly** to $f$ on a set $G$ if for every $z\in G$, $\forall\epsilon>0$, $\exists r>0$ such that for all $z\in B(z,r)$, $\forall n\geq N$, $|f_n(z)-f(z)|<\epsilon$.
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(choose $N$ based on $z$ and $\epsilon$)
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A sequence of functions $f_n$ **converges uniformly on compacta** to $f$ on a set $G$ if it converges uniformly on every compact subset of $G$.
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#### Theorem 5.?
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If the subsequence of a converging sequence of functions converges (a), then the original sequence converges (a).
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You can replace (a) with locally uniform convergence, uniform convergence, pointwise convergence, etc.
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#### UNKNOWN
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We defined $a^b=\{e^{b\log a}\}$ if $b$ is real, then $a^b$ is unique, if $b$ is complex, then $a^b=e^{b\log a}\{e^{2k\pi ik b}\},k\in\mathbb{Z}$.
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### Power series
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#### Definition 5.8
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A power series is a series of the form $\sum_{n=0}^{\infty}c_n(\zeta-\zeta_0)^n$.
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#### Theorem 5.10
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For every power series, there exists a radius of convergence $R$ such that the series converges absolutely and locally uniformly on $B(\zeta_0,R)$.
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And it diverges pointwise outside $B(\zeta_0,R)$.
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Proof:
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Without loss of generality, we can assume that $\zeta_0=0$.
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Suppose that the power series is $\sum_{n=0}^{\infty}c_n (\zeta)^n$ converges at $\zeta=re^{i\theta}$.
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We want to show that the series converges absolutely and uniformly on $\overline{B(0,r)}$ (_closed disk, I prefer to use this notation, although they use $\mathbb{D}$ for the disk (open disk)_).
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We know $c_n r^ne^{in\theta}\to 0$ as $n\to\infty$.
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So there exists $M\geq|c_n r^ne^{in\theta}|$ for all $n\in\mathbb{N}$.
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So $\forall \zeta\in\overline{B(0,r)}$, $|c_n\zeta^n|\leq |c_n|e^n\leq M(\frac{e}{r})^n$.
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So $\sum_{n=0}^{\infty}|c_n\zeta^n|$ converges absolutely.
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So the series converges absolutely and uniformly on $\overline{B(0,r)}$.
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_Some steps are omitted._
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EOP
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There are few cases for the convergence of the power series.
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Let $E=\{\zeta\in\mathbb{C}: \sum_{n=0}^{\infty}c_n(\zeta-\zeta_0)^n\text{ converges}\}$.
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1. It cloud only converge at $\zeta=0$.
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2. It could converge everywhere.
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Continue next time.
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