update notes

This commit is contained in:
Zheyuan Wu
2025-02-04 21:54:53 -06:00
parent 29ab09a045
commit b2b5496f2f
3 changed files with 152 additions and 1 deletions

View File

@@ -50,7 +50,7 @@ Which is in the form of circle equation.
EOP
## Chapter 4 Elements of functions
## Chapter 4 Elementary functions
> $e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}$

150
pages/Math416/Math416_L7.md Normal file
View File

@@ -0,0 +1,150 @@
# Lecture 7
## Review
### Exponential function
$$
e^z=e^{x+iy}=e^x(\cos y+i\sin y)
$$
### Logarithm
#### Definition 4.9 Logarithm
A logarithm of $a$ is any $b$ such that $e^b=a$.
### Branch of Logarithm
A branch of logarithm is a continuous function $f$ on a domain $D$ such that $e^{f(z)}=\exp(f(z))=z$ for all $z\in D$.
## Continue on Chapter 4 Elementary functions
### Logarithm
#### Theorem 4.11
$\log(\zeta)$ is holomorphic on $\mathbb{C}\setminus\{0\}$.
Proof:
We proved that $\frac{\partial}{\partial\overline{z}}e^{\zeta}=0$ on $\mathbb{C}\setminus\{0\}$.
Then $\frac{d}{dz}e^{\zeta}=\frac{\partial}{\partial x}e^{\zeta}=0$ if we know that $e^{\zeta}$ is holomorphic.
Since $\frac{d}{dz}e^{\zeta}=e^{\zeta}$, we know that $e^{\zeta}$ is conformal, so any branch of logarithm is also conformal.
Since $\exp(\log(\zeta))=\zeta$, we know that $\log(\zeta)$ is the inverse of $\exp(\zeta)$, so $\frac{d}{dz}\log(\zeta)=\frac{1}{e^{\log(\zeta)}}=\frac{1}{\zeta}$.
EOP
We call $\frac{f'}{f}$ the logarithmic derivative of $f$.
## Chapter 5. Power series
If $\sum_{n=0}^{\infty}c_n$ converges, then $\lim_{n\to\infty}c_n=0$ exists.
### Geometric series
$$
\sum_{n=0}^{N}c^n=\frac{1-c^{N+1}}{1-c}
$$
If $|c|<1$, then $\lim_{N\to\infty}\sum_{n=0}^{N}c^n=\frac{1}{1-c}$.
otherwise, the series diverges.
Proof:
The geometric series converges if $\frac{c^{N+1}}{1-c}$ converges.
$$
(1-c)(1+c+c^2+\cdots+c^N)=1-c^{N+1}
$$
If $|c|<1$, then $\lim_{N\to\infty}c^{N+1}=0$, so $\lim_{N\to\infty}(1-c)(1+c+c^2+\cdots+c^N)=1$.
If $|c|\geq 1$, then $c^{N+1}$ does not converge to 0, so the series diverges.
EOP
### Convergence
#### Definition 5.4
$$
\sum_{n=0}^{\infty}c_n
$$
converges absolutely if $\sum_{n=0}^{\infty}|c_n|$ converges.
Note: _Some other properties of converging series covered in Math4111, bad, very bad._
#### Definition 5.6 Convergence of sequence of functions
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 $n\geq N$, $|f_n(z)-f(z)|<\epsilon$.
(choose $N$ based on $z$)
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 $z\in G$, $|f_n(z)-f(z)|<\epsilon$.
(choose $N$ based on $\epsilon$)
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 $z\in B(z,r)$, $\forall n\geq N$, $|f_n(z)-f(z)|<\epsilon$.
(choose $N$ based on $z$ and $\epsilon$)
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 $G$.
#### Theorem 5.?
If the subsequence of a converging sequence of functions converges (a), then the original sequence converges (a).
You can replace (a) with locally uniform convergence, uniform convergence, pointwise convergence, etc.
#### UNKNOWN
We defined $a^b=\{e^{b\log a}\}$ if $b$ is real, then $a^b$ is unique, if $b$ is complex, then $a^b=e^{b\log a}\{e^{2k\pi ik b}\},k\in\mathbb{Z}$.
### Power series
#### Definition 5.8
A power series is a series of the form $\sum_{n=0}^{\infty}c_n(\zeta-\zeta_0)^n$.
#### Theorem 5.10
For every power series, there exists a radius of convergence $R$ such that the series converges absolutely and locally uniformly on $B(\zeta_0,R)$.
And it diverges pointwise outside $B(\zeta_0,R)$.
Proof:
Without loss of generality, we can assume that $\zeta_0=0$.
Suppose that the power series is $\sum_{n=0}^{\infty}c_n (\zeta)^n$ converges at $\zeta=re^{i\theta}$.
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)_).
We know $c_n r^ne^{in\theta}\to 0$ as $n\to\infty$.
So there exists $M\geq|c_n r^ne^{in\theta}|$ for all $n\in\mathbb{N}$.
So $\forall \zeta\in\overline{B(0,r)}$, $|c_n\zeta^n|\leq |c_n|e^n\leq M(\frac{e}{r})^n$.
So $\sum_{n=0}^{\infty}|c_n\zeta^n|$ converges absolutely.
So the series converges absolutely and uniformly on $\overline{B(0,r)}$.
_Some steps are omitted._
EOP
There are few cases for the convergence of the power series.
Let $E=\{\zeta\in\mathbb{C}: \sum_{n=0}^{\infty}c_n(\zeta-\zeta_0)^n\text{ converges}\}$.
1. It cloud only converge at $\zeta=0$.
2. It could converge everywhere.
Continue next time.

View File

@@ -9,4 +9,5 @@ export default {
Math416_L4: "Complex Variables (Lecture 4)",
Math416_L5: "Complex Variables (Lecture 5)",
Math416_L6: "Complex Variables (Lecture 6)",
Math416_L7: "Complex Variables (Lecture 7)",
}