update notations and fix typos

pages/CSE347/CSE347_L10.md

@@ -96,7 +96,7 @@ We wait for $R$ times and then take the stairs. In worst case, we wait for $R$ t
 
 Competitive ratio = $\frac{2R}{R}=2$.
 
-EOP
+QED
 
 Let's try $R=S-E$ instead.
 
@@ -116,7 +116,7 @@ We wait for $R=S-E$ times and then take the stairs.
 
 Competitive ratio = $\frac{S-E+S}{S}=2-\frac{E}{S}$.
 
-EOP
+QED
 
 What if we wait less time? Let's try $R=S-E-\epsilon$ for some $\epsilon>0$
 
@@ -174,7 +174,7 @@ The optimal offline solution: In each subsequence, must have at least $1$ miss.
 
 So the competitive ratio is at most $k+1$.
 
-EOP
+QED
 
 Using similar analysis, we can show that LRU is $k$ competitive.
 
@@ -184,7 +184,7 @@ Split the sequence into subsequences such that each subsequence LRU has $k$ miss
 
 Argue that OPT has at least $1$ miss in each subsequence.
 
-EOP
+QED
 
 #### Many sensible algorithms are $k$-competitive
 
@@ -210,7 +210,7 @@ So competitive ratio is at most $\frac{ck}{(c-1)k}=\frac{c}{c-1}$.
 
 _Actual competitive ratio is $\sim \frac{c}{c-1+\frac{1}{k}}$._
 
-EOP
+QED
 
 ### Conclusion
 
@@ -297,7 +297,7 @@ Let $P$ be a page in the cache with probability $1-\frac{1}{k}$.
 
 With probability $\frac{1}{k}$, $P$ is not in the cache and RAND evicts $P'$ in the cache and brings $P$ to the cache.
 
-EOP
+QED
 
 MRU is $k$-competitive.
 
@@ -317,4 +317,4 @@ Let's define the random variable $X$ as the number of misses of RAND MRU.
 
 $E[X]\leq 1+\frac{1}{k}$.
 
-EOP
+QED

pages/CSE347/CSE347_L5.md

@@ -161,7 +161,7 @@ $ISET(G,k)$  returns true if $G$ contains an independent set of size $\geq k$, a
 
 Algorithm? NO! We think that this is a hard problem.
 
-A lot of people have tried and could not find a poly-time solution
+A lot of pQEDle have tried and could not find a poly-time solution
 
 ### Example: Vertex Cover (VC)
 

pages/CSE347/CSE347_L7.md

@@ -154,7 +154,7 @@ This is a valid assignment since:
 - We pick either $v_i$ or $\overline{v_i}$
 - For each clause, at least one literal is true
 
-EOP
+QED
 
 Claim 2: If $\Psi$ is satisfiable, then Subset Sum has a solution.
 
@@ -174,7 +174,7 @@ Say $t=\sum$ elements we picked from $S$.
   - If $q_j=2$, then $z_j\in S'$
   - If $q_j=3$, then $y_j\in S'$
 
-EOP
+QED
 
 ### Example 2: 3 Color
 
@@ -228,13 +228,13 @@ For each dangler color is connected to blue, all literals cannot be blue.
 
 ...
 
-EOP
+QED
 
 Direction 2: If $G$ is 3-colorable, then $\Psi$ is satisfiable.
 
 Proof:
 
-EOP
+QED
 
 ### Example 3:Hamiltonian cycle problem (HAMCYCLE)
 

pages/CSE347/CSE347_L8.md

@@ -153,7 +153,7 @@ Summing over all vertices, the total number of crossing edges is at least $\frac
 
 So the total number of non-crossing edges is at most $\frac{|E|}{2}$.
 
-EOP
+QED
 
 #### Set cover
 
@@ -264,7 +264,7 @@ So $n(1-\frac{1}{k})^{|C|-1}=1$, $|C|\leq 1+k\ln n$.
 
 So the size of the set cover found is at most $(1+\ln n)k$.
 
-EOP
+QED
 
 So the greedy set cover is not too bad...
 
@@ -350,4 +350,4 @@ $$
 
 So the approximation ratio for greedy set cover is $H_d$.
 
-EOP
+QED

pages/CSE347/CSE347_L9.md

@@ -296,7 +296,7 @@ If $c'\geq 8c$, then $T(n)\leq c'n\log n+1$.
 
 $E[T(n)]\leq c'n\log n+1=O(n\log n)$
 
-EOP
+QED
 
 A more elegant proof:
 
@@ -345,5 +345,5 @@ E[X]&=\sum_{i=0}^{n-2}\sum_{j=i+1}^{n-1}\frac{2}{j-i+1}\\
 $$
 
 
-EOP
+QED
 

pages/CSE442T/CSE442T_L10.md

@@ -98,7 +98,7 @@ $x_1\equiv x_2\mod N$
 
 So it's one-to-one.
 
-EOP
+QED
 
 Let $y\in \mathbb{Z}_N^*$, letting $x=y^d\mod N$, where $d\equiv e^{-1}\mod \phi(N)$
 
@@ -130,7 +130,7 @@ By RSA assumption
 
 The second equality follows because for any finite $D$ and bijection $f:D\to D$, sampling $y\in D$ directly is equivalent to sampling $x\gets D$, then computing $y=f(x)$.
 
-EOP
+QED
 
 #### Theorem If inverting RSA is hard, then factoring is hard.
 

pages/CSE442T/CSE442T_L12.md

@@ -119,7 +119,7 @@ $\mathcal{D}$ can distinguish $x_{i+1}$ from a truly random $U_{i+1}$, knowing t
 
 So $\mathcal{D}$ can predict $x_{i+1}$ from $x_1\dots x_i$ (contradicting with that $X$ passes NBT)
 
-EOP
+QED
 
 ## Pseudorandom Generator
 

pages/CSE442T/CSE442T_L15.md

@@ -186,4 +186,4 @@ By hybrid argument, there exists a hybrid $H_i$ such that $D$ distinguishes $H_i
 
 For $H_0$, 
 
-EOP
+QED

pages/CSE442T/CSE442T_L16.md

@@ -35,7 +35,7 @@ $(r_1,F(r_1)),\ldots, (r_q,F(r_q))$
 
 So $D$ distinguishing output of $r_1,\ldots, r_q$ of PRF from the RF, this contradicts with definition of PRF.
 
-EOP
+QED
 
 Noe we have 
 

pages/CSE442T/CSE442T_L17.md

@@ -76,7 +76,7 @@ $$
 
 This contradicts the definition of hardcore bit.
 
-EOP
+QED
 
 ### Public key encryption scheme (multi-bit)
 
@@ -155,5 +155,5 @@ $$
 
 And proceed by contradiction. This contradicts the DDH assumption.
 
-EOP
+QED
 

pages/CSE442T/CSE442T_L20.md

@@ -72,7 +72,7 @@ So $\mathcal{B}$ can break the discrete log assumption with non-negligible proba
 
 So $h$ is a CRHF.
 
-EOP
+QED
 
 To compress by more, say $h_k:{0,1}^n\to \{0,1\}^{n-k},k\geq 1$, then we can use $h: \{0,1\}^{n+1}\to \{0,1\}^n$ multiple times.
 
@@ -119,7 +119,7 @@ Case 1: $h_i(m_1)=h_i(m_2)$, Then $\mathcal{A}$ finds a collision of $h$.
 
 Case 2: $h_i(m_1)\neq h_i(m_2)$, Then $\mathcal{A}$ produced valid signature on $h_i(m_2)$ after only seeing $Sign'_{sk'}(m_1)\neq Sign'_{sk'}(m_2)$. This contradicts the one-time secure of ($Gen,Sign,Ver$).
 
-EOP
+QED
 
 ### Many-time Secure Digital Signature
 

pages/CSE442T/CSE442T_L4.md

@@ -98,7 +98,7 @@ Proof:
 
     Then $P[a$ inverting $g]\sim P[a$ inverts $f$ all $q(n)]$ times. $<(1-\frac{1}{p(n)})^{q(n)}=(1-\frac{1}{p(n)})^{np(n)}<(e^{-\frac{1}{p(n)}})^{np(n)}=e^{-n}$ which is negligible function.
 
-EOP
+QED
 
 _we can always force the adversary to invert the weak one-way function for polynomial time to reach the property of strong one-way function_
 

pages/CSE559A/CSE559A_L13.md

@@ -2,6 +2,58 @@
 
 ## Positional Encodings
 
+### Fixed Positional Encodings
+
+Set of sinusoids of different frequencies.
+
+$$
+f(p,2i)=\sin(\frac{p}{10000^{2i/d}})\quad f(p,2i+1)=\cos(\frac{p}{10000^{2i/d}})\
+$$
+
+[source](https://kazemnejad.com/blog/transformer_architecture_positional_encoding/)
+
+### Positional Encodings in Reconstruction
+
+MLP is hard to learn high-frequency information from scaler input $(x,y)$.
+
+Example: network mapping from $(x,y)$ to $(r,g,b)$.
+
+### Generalized Positional Encodings
+
+- Dependence on location, scaler, metadata, etc.
+- Can just be fully learned (use `nn.Embedding` and optimize based on a categorical input.)
+
 ## Vision Transformer (ViT)
 
+### Class Token
+
+In Vision Transformers, a special token called the class token is added to the input sequence to aggregate information for classification tasks.
+
+### Hidden CNN Modules
+
+- PxP convolution with stride P (split the image into patches and use positional encoding)
+
+### ViT + ResNet Hybrid
+
+Build a hybrid model that combines the vision transformer after 50 layer of ResNet.
+
 ## Moving Forward
+
+At least for now, CNN and ViT architectures have similar performance at least in ImageNet.
+
+- General Consensus: once the architecture is big enough, and not designed terribly, it can do well.
+- Differences remain:
+  - Computational efficiency
+  - Ease of use in other tasks and with other input data
+  - Ease of training
+
+## Wrap up
+
+Self attention as a key building block
+
+Flexible input specification using tokens with positional encodings
+
+A wide variety of architectural styles
+
+Up Next:
+Training deep neural networks

pages/CSE559A/mlp_image_reconstruction.py

@@ -0,0 +1,82 @@
+import torch
+from torchvision import transforms
+from PIL import Image
+import matplotlib.pyplot as plt
+
+class MLPScalar(torch.nn.Module):
+    # Define your MLPScalar architecture here
+
+    def __init__(self):
+        super(MLPScalar, self).__init__()
+        # Example architecture
+        self.fc1 = torch.nn.Linear(2, 128)
+        self.fc2 = torch.nn.Linear(128, 3)  # Outputs RGB
+
+    def forward(self, x):
+        x = torch.nn.functional.relu(self.fc1(x))
+        x = torch.sigmoid(self.fc2(x))  # Normalize output to [0, 1]
+        return x
+
+class MLPPositional(torch.nn.Module):
+    # Define your MLPPositional architecture here
+
+    def __init__(self, num_frequencies=10, include_input=True):
+        super(MLPPositional, self).__init__()
+        # Example architecture
+        self.include_input = include_input
+        self.fc1 = torch.nn.Linear(2, 128)
+        self.fc2 = torch.nn.Linear(128, 3)  # Outputs RGB
+
+    def forward(self, x):
+        if self.include_input:
+            # Process coordinates, add positional encoding here if needed
+            x = torch.cat([x, self.positional_encoding(x)], dim=-1)
+        x = torch.nn.functional.relu(self.fc1(x))
+        x = torch.sigmoid(self.fc2(x))  # Normalize output to [0, 1]
+        return x
+
+    def positional_encoding(self, x):
+        # Example positional encoding
+        return torch.cat([torch.sin(x * (2 ** i)) for i in range(10)], dim=-1)
+
+if __name__ == '__main__':
+    # Load a real image
+    image_path = input()[1:-1]  # Replace with your image file path
+    image = Image.open(image_path).convert('RGB')
+    
+    # Normalize and resize the image
+    transform = transforms.Compose([
+        transforms.Resize((256, 256)),  # Resize image to desired dimensions
+        transforms.ToTensor(),           # Convert to Tensor and normalize to [0,1]
+    ])
+    
+    image_tensor = transform(image)
+    
+    # Create dummy normalized coordinates (assume image coordinates normalized to [0,1])
+    coords = torch.rand(10, 2)  # 10 random coordinate pairs
+    print("Input coordinates:")
+    print(coords)
+
+    # Test MLP with scalar input
+    model_scalar = MLPScalar()
+    out_scalar = model_scalar(coords)
+    print("\nMLPScalar output (RGB):")
+    print(out_scalar)
+
+    # Test MLP with positional encoding
+    model_positional = MLPPositional(num_frequencies=10, include_input=True)
+    out_positional = model_positional(coords)
+    print("\nMLPPositional output (RGB):")
+    print(out_positional)
+
+    # Optionally, use the output to create a new image
+    output_image = (out_positional.view(10, 1, 3) * 255).byte().numpy()  # Reshape and scale
+    output_image = output_image.transpose(0, 2, 1)  # Prepare for visualization
+
+    # Visualize the output
+    plt.figure(figsize=(10, 2))
+    for i in range(output_image.shape[0]):
+        plt.subplot(2, 5, i + 1)
+        plt.imshow(output_image[i].reshape(1, 3), aspect='auto')
+        plt.axis('off')
+    plt.show()

pages/Math4111/Exam_reviews/Math4111_E2.md

@@ -22,7 +22,7 @@ Suppose $S$ and $T$ are compact, let $\{G_\alpha\}_{\alpha\in A}$ be an open cov
 
 ...
 
-EOP
+QED
 
 ## K-cells are compact
 
@@ -63,7 +63,7 @@ Let $n\in \mathbb{N}$ be such that $\frac{1}{2^n}<r$. Then by $(c)$, $I(n)\subse
 
 Then $\{G_{\alpha_0}\}$ is a cover of $I_n$ which contradicts with (b)
 
-EOP
+QED
 
 ## Redundant subcover question
 

pages/Math4111/Math4111_L1.md

@@ -36,7 +36,7 @@ So $m^2$ is divisible by 4, $2n^2$ is divisible by 4.
 
 So $n^2$ is even. but they are not both even.
 
-EOP
+QED
 
 ### Theorem (No closest rational for a irrational number)
 

pages/Math4111/Math4111_L10.md

@@ -47,7 +47,7 @@ $\impliedby$ Suppose $K$ is compact relative to $Y$. Let $\{G_\alpha\}_{\alpha\i
 
 Since $k$ is compact relative to $Y$, $\{G_\alpha\cap Y\}_\alpha$ has a finite subcover $\{G_{\alpha_i}\cap Y\}_{i=1}^n$. Then $\{G_{\alpha_i}\}_{i=1}^n$ is a finite subcover of $\{G_\alpha\}_{\alpha}$ of $K$.
 
-EOP
+QED
 
 #### Theorem 2.24
 

pages/Math4111/Math4111_L11.md

@@ -31,7 +31,7 @@ And $(\bigcup_{\alpha\in A} G_\alpha)\cup F^c=X\supset K$ is an open cover of $K
 
 Since $F\subset K$, So $\Phi$ is a cover of $F$ then $\Phi\backslash \{F^c\}$ is a finite subcover of $\{G_\alpha\}_{\alpha \in A}$ of $F$.
 
-EOP
+QED
 
 ##### Corollary 2.35
 
@@ -57,7 +57,7 @@ Let $l=\max\{n_1,...,n_m\}$, then $K_l=\bigcap^{m}_{i=i} K_{n_i}$. So $K_1\subse
 
 which contradicts with $K_l\subset K_1$
 
-EOP
+QED
 
 #### Theorem 2.37
 
@@ -75,7 +75,7 @@ For each $q\in K, q\notin E'$, so $\exists$ neighborhood $\forall q\in B_{r_q}(q
 
 Then $\{V_q\}_{q\in K}$ is an open cover of $K$ , so it has a finite subcover $\{V_{q_i}\}^n_{n=1}$. Then $E\subset K\subset \bigcup_{i=1}^n$, and $\forall i,V_{q_i}\cap E\subset \{q_i\}$, then $E\subset\{q_1,...,q_n\}$
 
-EOP
+QED
 
 #### Theorem 2.38
 
@@ -101,4 +101,4 @@ Since $x$ is an upper bound of $E$ and $a_n\in E$, then $a_n\leq x$.
 
 Since $x$ is the least upper bound of $E$, and $b_n$ is an upper bound of $E$, then $x\leq b_n$. $x\in E,E\neq \phi$
 
-EOP
+QED

pages/Math4111/Math4111_L12.md

@@ -20,7 +20,7 @@ Proof:
 
 Fix $p\in X$, then $\{B_n(p)\}_{n\in \mathbb{N}}$ (specific open cover) is an open cover of $S$ (Since $\bigcup_{n\in \mathbb{N}}=X$). Since $S$ is compact, then $\exists$ a finite subcover ${n\in \mathbb{N}}_{i=1}^k=S$, let $r=max(n_1,...n_k)$, Then $S\subset B_r(p)$
 
-EOP
+QED
 
 #### Definition k-cell
 
@@ -74,7 +74,7 @@ Let $n\in \mathbb{N}$ be such that $\frac{1}{2^n}<r$. Then by $(c)$, $I(n)\subse
 
 Then $\{G_{\alpha_0}\}$ is a cover of $I_n$ which contradicts with (b)
 
-EOP
+QED
 
 #### Theorem 2.41
 
@@ -131,4 +131,4 @@ $$
 
 So $B_r(y)\cap S$ is finite. By **Theorem 2.20**, $y\notin  S$, this proves the claim so $S'\cap E=\phi$
 
-EOP
+QED

pages/Math4111/Math4111_L13.md

@@ -90,7 +90,7 @@ Thus, $\exists r>0$ such that $(w-r,w+r)\cap B=\phi$.
 
 Let $z=w+\frac{r}{2}$, then $x,y,z$ satisfy the desired properties.
 
-EOP
+QED
 
 ## Chapter 3: Numerical Sequences and Series
 
@@ -140,4 +140,4 @@ Let $n\geq N$ (arbitrary)
 
 Then $|s_n-q|=\frac{1}{n}\leq \frac{1}{N}\leq \epsilon$
 
-EOP
+QED

pages/Math4111/Math4111_L14.md

@@ -74,7 +74,7 @@ And $\forall \epsilon>0,d(p,p')<2\epsilon\implies d(p,p')=0$. So $p=p'$
 
 Let $\epsilon>0$. Choose $N\in \mathbb{N}$ such that $N>\frac{1}{\epsilon}$. Then if $n\geq N$, $d(p_n,p)<\frac{1}{n}\leq \frac{1}{N}\leq \epsilon$
 
-EOP
+QED
 
 #### Theorem 3.3
 

pages/Math4111/Math4111_L15.md

@@ -14,7 +14,7 @@ $$
     Proof:  
     Let $\epsilon>0$ be arbitrary, then $\exists N$ such that $\forall n\geq N,|a_n-a|<\epsilon$.
     Then if $n\geq N$, $|a_n-a|\leq \sqrt{|a_n-a|^2}\leq\sqrt{|a_n-a|^2+|b_n-b|^2}=|x_n-(a,b)|<\epsilon$.  
-    EOP  
+    QED  
     (b) If $x_n\to (a,b)$, then $b_n\to b$.
     This follows from the same argument from (a)
 2. Prove the $\implies$ direction.
@@ -28,7 +28,7 @@ $$
     Since $a_n\to a$, $\exists N_1$ such that $\forall n\geq N_1,|a_n-a|<\frac{\epsilon}{\sqrt{2}}$.  
     Since $b_n\to b$, $\exists N_2$ such that $\forall n\geq N_2,|b_n-b|<\frac{\epsilon}{\sqrt{2}}$.  
     Let $N=\max\{N_1,N_2\}$. Then if $n\geq N$, $|a_n-a|<\epsilon$ and $|b_n-b|<\sqrt{\frac{\epsilon^2}{2}+\frac{\epsilon^2}{2}}=\epsilon$.  
-    EOP  
+    QED  
 
 ## New Materials
 
@@ -82,7 +82,7 @@ $$
 \left|\frac{1}{s_n}-\frac{1}{s}\right|=\frac{|s-s_n|}{|s||s_n|}<\frac{\frac{\epsilon|s|^2}{2}}{|s|^2}=\epsilon
 $$
 
-EOP
+QED
 
 ### Subsequences
 
@@ -108,7 +108,7 @@ $\implies$:
 
 Thought process: show what if the sequence does not converge to $p$, then there exists a subsequence that does not converge to $p$.
 
-EOP
+QED
 
 #### Theorem 3.6
 

pages/Math4111/Math4111_L16.md

@@ -17,7 +17,7 @@ Let $n\geq N$. Since $(s_n)$ is monotonic increasing, $s_n\geq s_N>t-\epsilon$.
 
 So $(s_n)$ converges to $t$.
 
-EOP
+QED
 
 ## New materials
 
@@ -47,7 +47,7 @@ Step 3: By induction, we can get a sequence $n_1,n_2,\cdots$ such that $\forall
 
 Then $(p_{n_i})$ is a subsequence of $(p_n)$ and $p_{n_i}\to q$.
 
-EOP
+QED
 
 ### Cauchy Sequences
 
@@ -93,7 +93,7 @@ If $m,n\geq N$, then $d(p_m,p_n)\leq d(p_m,p)+d(p,p_n)<\epsilon+\epsilon=2\epsil
 
 *You can also use $\frac{\epsilon}{2}$ instead of $\epsilon$ in the above proof, just for fun.*
 
-EOP
+QED
 
 #### Lemma 3.11 (b)
 
@@ -107,7 +107,7 @@ Let $r=max\{d(p_i,p_j);1\leq i,j\leq N\}+1$.
 
 Then $\forall n\in \mathbb{N}$, $p_n\in B_r(p_N)$.
 
-EOP
+QED
 
 > Note: This proof is nearly identical to the proof of convergent sequences implies bounded.
 
@@ -148,5 +148,5 @@ By definition of supremum, the claim implies that $\forall \epsilon>0$, $diam(\o
 
 (b) By **Theorem 2.36**, $\bigcap_{n=1}^{\infty}K_n\neq \phi$. Suppose for contradiction that there are at least two distinct points $p,q\in \bigcap_{n=1}^{\infty}K_n$. Then for all $n\in \mathbb{N}$, $x,y\in K_n$ so $diam K_n\geq d(p,q)>0$. Then diameter of $K_n$ does not converge to 0.
 
-EOP
+QED
 

pages/Math4111/Math4111_L17.md

@@ -44,7 +44,7 @@ Note that **Theorem 2.41** only works for $\mathbb{R}^k$.
 
 So by (b), $(p_n)$ converges to some $p\in \overline{B(0,R)}$.
 
-EOP
+QED
 
 #### Definition 3.12
 
@@ -89,7 +89,7 @@ If $(s_n)$ is monotonic and bounded, then by previous result, $(s_n)$ converges.
 
 If $(s_n)$ is monotonic and converges, then by **Theorem 3.2(c)**, $(s_n)$ is bounded.
 
-EOP
+QED
 
 ### Upper and lower limits
 

pages/Math4111/Math4111_L18.md

@@ -79,7 +79,7 @@ Case 2: $(s_n)$ is bounded above.
 
 Let $M$ be an upper bound for $(s_n)$. Then $\{n\in\mathbb{N}:s_n\in[M,x]\}$ is infinite, by **Theorem 3.6 (b)** ($\exists$ subsequence $(s_{n_k})$ in $[x,M)$ and $\exists t\in[x,M]$ such that $s_{n_k}\to t$. This implies $t\in E$, so $x\leq t\leq S^*$).
 
-EOP
+QED
 
 #### Theorem 3.19 ("one-sided squeeze theorem")
 
@@ -104,7 +104,7 @@ By **Theorem 3.17**, $\{n\in\mathbb{N}:t_n\geq x\}$ is finite $\implies$ $\{n\in
 
 Thus $\limsup_{n\to\infty} s_n\leq \limsup_{n\to\infty} t_n$.
 
-EOP
+QED
 
 > Normal squeeze theorem: If $s_n\leq t_n\leq u_n$ for all $n\in\mathbb{N}$, and $\lim_{n\to\infty} s_n=\lim_{n\to\infty} u_n=L$, then $\lim_{n\to\infty} t_n=L$.
 >

pages/Math4111/Math4111_L19.md

@@ -15,14 +15,14 @@
     &\geq\binom{n}{4}
     \end{aligned}
     $$
-    EOP
+    QED
 2. Using part 1, show that $\lim_{n\to\infty}\frac{n^3}{2^n}=0$.  
    Proof:
    $$
    \frac{n^3}{2^n}\leq\frac{n^3}{\binom{n}{4}}
    $$
    The value of $\frac{n^3}{\binom{n}{4}}$ is decreasing when $n\geq 4$.
-   EOP
+   QED
 
 ## New materials
 
@@ -53,7 +53,7 @@ $$
 |s_m-s_n|=\left|\sum_{k=n}^{m}a_k\right|<\epsilon.
 $$
 
-EOP
+QED
 
 Special case of this theorem.
 
@@ -94,7 +94,7 @@ $$
 \left|\sum_{k=n}^{m}c_k\right|\leq \sum_{k=n}^{m}c_k<\epsilon.
 $$
 
-EOP
+QED
 
 #### Theorem 3.26 (Geometric series)
 
@@ -115,7 +115,7 @@ So $s_n=\frac{1-x^{n+1}}{1-x}$.
 
 Since $|x|<1$, $x^{n+1}$ converges to 0. So $\lim_{n\to\infty}s_n=\frac{1}{1-x}$.
 
-EOP
+QED
 
 #### Lemma 3.28 
 
@@ -149,7 +149,7 @@ $$
 
 > Fun fact: $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$.
 
-EOP
+QED
 
 #### Theorem 3.27 (Cauchy condensation test)
 
@@ -191,4 +191,4 @@ So $(s_n)_{n=1}^{\infty}$ is a bounded above.
 
 By **Theorem 3.14**, $(s_n)_{n=1}^{\infty}$ converges if and only if $(t_k)_{k=0}^{\infty}$ converges.
 
-EOP
+QED

pages/Math4111/Math4111_L2.md

@@ -59,7 +59,7 @@ Suppose for contradiction $\alpha$ and $\beta$ are both LUB of $E$, then $\alpha
 
 WLOG $\alpha>\beta$ and $\beta>\alpha$.
 
-EOP
+QED
 
 We write $\sup E$ to denote the LUB of $E$.
 

pages/Math4111/Math4111_L20.md

@@ -114,7 +114,7 @@ Therefore, $e\leq \liminf_{n\to\infty} t_n\leq \limsup_{n\to\infty} t_n\leq e$.
 
 So $\lim_{n\to\infty} t_n$ exists and $\lim_{n\to\infty} t_n = e$.
 
-EOP
+QED
 
 #### Theorem 3.32
 
@@ -156,7 +156,7 @@ $$
 
 Contradiction.
 
-EOP
+QED
 
 ### The root and ratio tests
 
@@ -190,7 +190,7 @@ Thus $a_n\not\to 0$, $\sum_{n=0}^{\infty} a_n$ diverges.
 
 (c) $\sum_{n=0}^{\infty} \frac{1}{n}$ and $\sum_{n=0}^{\infty} \frac{1}{n^2}$ both have $\alpha = 1$. but the first diverges and the second converges.
 
-EOP
+QED
 
 #### Theorem 3.34 (Ratio test)
 
@@ -232,7 +232,7 @@ i.e. $\forall n\geq N, |a_n| < \beta^{n-N}|a_N|=\beta^n(\beta^{-N}|a_N|)$.
 
 Since $\sum_{n=N}^{\infty} \beta^n$ converges, by comparison test, $\sum_{n=0}^{\infty} a_n$ converges.
 
-EOP
+QED
 
 We will skip **Theorem 3.37**. One implication is that if ratio test can be applied, then root test can be applied.
 
@@ -267,4 +267,4 @@ $$
 
 By root test, the series converges absolutely for all $z\in\mathbb{C}$ with $|z| < R$.
 
-EOP
+QED

pages/Math4111/Math4111_L21.md

@@ -56,7 +56,7 @@ $$
 \end{aligned}
 $$
 
-EOP
+QED
 
 #### Theorem 3.42 (Dirichlet's test)
 
@@ -101,7 +101,7 @@ $$
 
 So $\sum a_nb_n$ converges.
 
-EOP
+QED
 
 #### Theorem 3.43 (Alternating series test)
 
@@ -122,7 +122,7 @@ So $|A_n|\leq 1$ for all $n\in \mathbb{N}$.
 
 By Theorem 3.42, $\sum_{n=1}^\infty a_n b_n$ converges.
 
-EOP
+QED
 
 Example:
 
@@ -161,7 +161,7 @@ So $|A_n|\leq \frac{2}{|1-z|}$ for all $n\in \mathbb{N}$.
 
 By Dirichlet's test, $\sum_{n=0}^\infty b_nz^n$ 
 
-EOP
+QED
 
 ### Absolute convergence
 
@@ -183,7 +183,7 @@ $$
 \sum_{n=0}^\infty |a_n|\geq \sum_{n=0}^\infty a_n
 $$
 
-EOP
+QED
 
 Rearrangement of series:
 
@@ -257,4 +257,4 @@ For every $n\in \mathbb{N}$, there exists a $p$ such that $\{1,2,\cdots,n\}\subs
 
 Then $|s_n-t_n|\leq \sum_{k=N+1}^\infty |a_k|$.
 
-EOP
+QED

pages/Math4111/Math4111_L22.md

@@ -77,7 +77,7 @@ This proves that $\lim_{n\to\infty} |s_n - t_n| = 0$.
 
 Since $\lim_{n\to\infty} s_n$ exists, $\lim_{n\to\infty} s_n = \lim_{n\to\infty} t_n$.
 
-EOP
+QED
 
 #### Theorem 3.54
 
@@ -137,7 +137,7 @@ Then: $(p_n)$ is a sequence in $E\backslash\{p\}$ with $d_X(p_n,p) = \frac{1}{n}
 
 So $\lim_{n\to\infty} f(p_n) \neq q$.
 
-EOP
+QED
 
 With this theorem, we can use the properties of limit of sequences to study limits of functions.
 

pages/Math4111/Math4111_L24.md

@@ -33,7 +33,7 @@ $\impliedby$: Suppose for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X
 
 Since $p\in f^{-1}(B_\epsilon(f(p)))$ and $f^{-1}(B_\epsilon(f(p)))$ is open, $\exists \delta > 0$ such that $B_\delta(p)\subset f^{-1}(B_\epsilon(f(p)))$. Therefore, $f(B_\delta(p))\subset B_\epsilon(f(p))$. This shows that $f$ is continuous.
 
-EOP
+QED
 
 #### Corollary 4.8
 
@@ -82,7 +82,7 @@ Proof:
 
 Let $p\in \mathbb{R}$ and $\epsilon > 0$. Let $\delta = \epsilon$. Then, $\forall x\in \mathbb{R}$, if $|x-p|<\delta$, then $|f(x)-f(p)| = |x-p| < \delta = \epsilon$.
 
-EOP
+QED
 
 Therefore, by **Theorem 4.9**, $f(x) = x^2$ is continuous. $f(x) = x^3$ is continuous... So all polynomials are continuous.
 
@@ -131,7 +131,7 @@ By Theorem 2.41, $f(X)$ is closed and bounded.
 
 By Theorem 2.28, $\sup f(X)$ and $\inf f(X)$ exist and are in $f(X)$. Let $p_0\in X$ such that $f(p_0) = \sup f(X)$. Let $q_0\in X$ such that $f(q_0) = \inf f(X)$.
 
-EOP
+QED
 
 ---
 
@@ -151,7 +151,7 @@ Proof:
 
 See the textbook.
 
-EOP
+QED
 
 ---
 
@@ -191,7 +191,7 @@ Since $A$ and $B$ are separated, $\overline{A}\cap B = \phi$ and $\overline{B}\c
 
 Therefore, $\overline{G}\cap H = \phi$ and $\overline{H}\cap G = \phi$.
 
-EOP
+QED
 
 #### Theorem 4.23 (Intermediate Value Theorem)
 
@@ -207,4 +207,4 @@ $f(a)$ and $f(b)$ are real numbers in $f([a,b])$, and $c$ is a real number betwe
 
 By **Theorem 2.47**, $c\in f([a,b])$.
 
-EOP
+QED

pages/Math4111/Math4111_L25.md

@@ -99,7 +99,7 @@ _The $\epsilon$ bound would not hold if we only had pointwise convergence._
 
 $|f_N(x) - f_N(p)| < 3\epsilon$.
 
-EOP
+QED
 
 > Recall: If $(s_n)$ is a sequence in $\mathbb{R}$, then $(s_n)$ converges to $s$ if and only if it is Cauchy.  
 > i.e. $\forall \epsilon > 0$, $\exists N\in \mathbb{N}$, $\forall n, m\geq N$, $|s_n - s_m| < \epsilon$.
@@ -147,7 +147,7 @@ $$
 \end{aligned}
 $$
 
-EOP
+QED
 
 Example:
 

pages/Math4111/Math4111_L3.md

@@ -22,10 +22,12 @@ Let $S=\mathbb{Z}$.
 
 ## Continue
 
-### LUBP
+### LUBP (The least upper bound property)
 
 Proof that $LUBP\implies GLBP$.
 
+Proof:
+
 Let $S$ be an ordered set with LUBP. Let $B<S$ be non-empty and bounded below.
 
 Let $L=y\in S:y$ is a lower bound of $B$. From the picture, we expect $\sup L=\inf B$ First we'll show $\sup L$ exists.
@@ -55,6 +57,8 @@ Let's say $\alpha=sup\ L$. We claim that $\alpha=inf\ B$. We need to show $2$ th
 
 Thus $\alpha=inf\ B$
 
+QED
+
 ### Field
 
 |                | addition                                                    | multiplication                                                 |

pages/Math4111/Math4111_L4.md

@@ -39,7 +39,7 @@ This implies $(m+1)x>\alpha$
 
 Since $(m+1)x\in \alpha$, this contradicts the fact that $\alpha$ is an upper bound of $A$.
 
-EOP
+QED
 
 ### $\mathbb{Q}$ is dense in $\mathbb{R}$
 
@@ -59,7 +59,7 @@ By Archimedean property, $\exist n\in \mathbb{N}$ such that $n(y-x)>1$, and $\ex
 
 So $-m_2<nx<m_1$. Thus $\exist m\in \mathbb{Z}$ such that $m-1\leq nx<m$ (Here we use a property of $\mathbb{Z}$) We have $ny>1+nx\geq 1+(m-1)=m$
 
-EOP
+QED
 
 ### $\sqrt{2}\in \mathbb{R}$, $(\sqrt[n]{x}\in\mathbb{R})$
 

pages/Math4111/Math4111_L5.md

@@ -94,6 +94,8 @@ So want $k\leq \frac{y^n-x}{ny^{n-1}}$
 
 [For actual proof, see the text.]
 
+QED
+
 ### Complex numbers
 
 1. $=\{a+bi:a,b\in \mathbb{R}\}$.
@@ -149,7 +151,9 @@ $$
 (\sum a_j b_j)^2=(\sum a_j^2)(\sum b_j^2)
 $$
 
-Proof for real numbers:
+Proof:
+
+For real numbers:
 
 Let $A=\sum a_j^2,B=\sum b_j^2, C=\sum a_j b_j$, want to show $C^2\leq AB$
 
@@ -165,6 +169,8 @@ let $t=C/B$ to get $0\leq A-2(C/B)C+(C/B)^2B=A-\frac{C^2}{B}$
 
 to generalize this to $\mathbb{C}$, $A=\sum |a_j|^2,B=\sum |b_j|^2,C=\sum |a_j \bar{b_j}|$.
 
+QED
+
 ### Euclidean spaces
 
 Nothing much to say. lol.

pages/Math4111/Math4111_L6.md

@@ -126,7 +126,9 @@ $A$ is countable, $n\in \mathbb{N}$,
 
 $\implies A^n=\{(a_{1},...,a_{n}):a_1\in A, a_n\in A\}$, is countable.
 
-Proof: Induct on $n$, 
+Proof:
+
+Induct on $n$,
 
 Base case $n=1$,
 
@@ -136,14 +138,20 @@ Induction step: suppose $A^{n-1}$ is countable. Note $A^n=\{(b,a):b\in A^{n-1},a
 
 Since $b$ is fixed, so this is in 1-1 correspondence with $A$, so it's countable by Theorem 2.12.
 
+QED
+
 #### Theorem 2.14
 
 Let $A$ be the set of all sequences for 0s and 1s. Then $A$ is uncountable.
 
-Proof: Let $E\subset A$ be a countable subset. We'll show $A\backslash E\neq \phi$ (i.e.$\exists t\in A$ such that $t\notin E$)
+Proof:
+
+Let $E\subset A$ be a countable subset. We'll show $A\backslash E\neq \phi$ (i.e.$\exists t\in A$ such that $t\notin E$)
 
 $E$ is countable so we can list it's elements $S_1,S_2,S_3,...$.
 
 Then we define a new sequence $t$ which differs from $S_1$'s first bit and $S_2$'s second bit,...
 
 This is called Cantor's diagonal argument.
+
+QED

pages/Math4111/Math4111_L7.md

@@ -80,12 +80,16 @@ Let $(X,d)$ be  a metric space, $\forall p\in X,\forall r>0$, $B_r(p)$ is an ope
 
 *every ball is an open set*
 
-Proof: Let $q\in B_r(p)$.
+Proof:
+
+Let $q\in B_r(p)$.
 
 Let $h=r-d(p,q)$.
 
 Since $q\in B_r(p),h>0$. We claim that $B_h(q)$. Then $d(q,s)<h$, so $d(p,s)\leq d(p,q)+d(q,s)<d(p,q)+h=r$. (using triangle inequality) So $S\in B_r(p)$.
 
+QED
+
 ### Closed sets
 
 1. $E\subset X,p\in X$. We say $p$ is a limit point of $E$ if $\forall r>0, (B_r(p)\cap E)\backslash {p}\neq \phi$.
@@ -94,7 +98,9 @@ Since $q\in B_r(p),h>0$. We claim that $B_h(q)$. Then $d(q,s)<h$, so $d(p,s)\leq
 
 2. $E$ is closed if $E'\subset E$
 
-Example: $X=\mathbb{R}^2$, $E=[0,1)\times [0,1)$.
+Example:
+
+$X=\mathbb{R}^2$, $E=[0,1)\times [0,1)$.
 
 $(1,1)$ is a limit point.
 

pages/Math4111/Math4111_L8.md

@@ -41,7 +41,7 @@ let $B_s(p)\cap E)\backslash \{p\}={q_1,...,q_n}$
 
 Then $(B_s(p)\cap E)\backslash \{p\}=\phi$, so $p\notin E$
 
-EOP
+QED
 
 #### Theorem 2.22 De Morgan's law
 
@@ -95,7 +95,7 @@ $$
 
 So $(E^c)'\subset E^c$
 
-EOP
+QED
 
 #### Theorem 2.24
 
@@ -105,7 +105,7 @@ Proof:
 
 Suppose $\forall \alpha, G_\alpha$ is open. Let $x\in \bigcup _{\alpha} G_\alpha$. Then $\exists \alpha_0$ such that $x\in G_{\alpha_0}$. Since $G_{\alpha_0}$ is open, $\exists r>0$ such that $B_r(x)\subset G_{\alpha_0}$ Then $B_r(x)\subset G_{\alpha_0}\subset \bigcup_{\alpha} G_\alpha$
 
-EOP
+QED
 
 ##### A finite intersection of open set is open
 
@@ -117,7 +117,7 @@ Let $x\in \bigcap^n_{i=1}G_i$, then $\forall i\in \{1,..,n\}$ and $G_i$ is open,
 
 Let $r=min\{r_1,...,r_n\}$. Then $\forall i\in \{1,...,n\}$. $B_r(x)\subset B_{r_i}(x)\subset G_i$. So $B_r(x)\subset \bigcup_{i=1}^n G_i$
 
-EOP
+QED
 
 The other two can be proved by **Theorem 2.22,2.23**
 
@@ -147,4 +147,4 @@ This proves (b)
 
 So $\bar{E}^c$ is open
 
-EOP
+QED

pages/Math4111/Math4111_L9.md

@@ -37,7 +37,7 @@ Let $h>0$. Since $y-h$ is not an upper bound of $E$, $\exists x\in E$ such that
 
 Since $y$ is an upper bound of $E$, $x\leq y$. So $x\in B_n(y)\cap E$, so $B_h(y)\cap E\neq \phi$.
 
-EOP
+QED
 
 #### Remark 2.29
 
@@ -75,7 +75,7 @@ To show $G\cap Y\subset E$.
 
 $G\cap Y=\left(\bigcup_{p\in E}V_p\right)\cap Y=\bigcup_{p\in E}(V_p\cap Y)\subset E$
 
-EOP
+QED
 
 ### Compact sets
 
@@ -101,4 +101,4 @@ as we can build an infinite open cover $\bigcup_{i\in Z} (i,i+2)$ and it does no
 
 Suppose we consider the sub-collection $\{n_i,n_i+2:i=1,..,k\}$, Then $N+3$ is not in the union, where $N=max\{n_1,...,n_k\}$.
 
-EOP
+QED

pages/Math4111/index.md

@@ -10,7 +10,7 @@ Topics include:
 4. Convergence of Series and Sequences
 5. Limits and Continuity
 
-The course is taught by [Alan Chang](https://math.wustl.edu/people/alan-chang).
+The course is taught by [Alan Chang](https://math.wustl.edu/pQEDle/alan-chang).
 
 It is easy in my semester perhaps, it is the first course I got 3 perfect scores in exams. (Unfortunately, I did not get the extra credit for the third midterm exam.)
 

pages/Math4121/Math4121_L10.md

@@ -40,7 +40,7 @@ By **Theorem 6.11**, $f^2,g^2\in \mathscr{R}(\alpha)$ on $[a, b]$.
 
 By linearity, $fg=1/4((f+g)^2-(f-g)^2)\in \mathscr{R}(\alpha)$ on $[a, b]$.
 
-EOP
+QED
 
 (b) $|f|\in \mathscr{R}(\alpha)$ on $[a, b]$, and $|\int_a^b f d\alpha|\leq \int_a^b |f| d\alpha$.
 
@@ -54,7 +54,7 @@ Let $c=-1$ or $c=1$. such that $c\int_a^b f d\alpha=| \int_a^b f d\alpha|$.
 
 By linearity, $c\int_a^b f d\alpha=\int_a^b cfd\alpha$. Since $cf\leq |f|$, by monotonicity, $|\int_a^b cfd\alpha|=\int_a^b cfd\alpha\leq \int_a^b |f| d\alpha$.
 
-EOP
+QED
 
 ### Indicator Function
 
@@ -104,6 +104,6 @@ Since $f$ is continuous at $s$, when $x\to s$, $U(P,f,\alpha)\to f(s)$ and $L(P,
 
 Therefore, $U(P,f,\alpha)-L(P,f,\alpha)\to 0$, $f\in \mathscr{R}(\alpha)$ on $[a, b]$, and $\int_a^b f d\alpha=f(s)$.
 
-EOP
+QED
 
 

pages/Math4121/Math4121_L11.md

@@ -79,7 +79,7 @@ $$
 \end{aligned}
 $$
 
-EOP
+QED
 
 If $f\in \mathscr{R}$, and there exists a differentiable function $F:[a,b]\to \mathbb{R}$ such that $F'=f$ on $(a,b)$, then
 
@@ -107,4 +107,4 @@ $$
 
 So, $f\in \mathscr{R}$ and $\int_a^b f(x)\ dx=F(b)-F(a)$.
 
-EOP
+QED

pages/Math4121/Math4121_L14.md

@@ -26,7 +26,7 @@ Proof:
 
 To prove Riemann's Integrability Criterion, we need to show that a bounded function $f$ is Riemann integrable if and only if for every $\sigma, \epsilon > 0$, there exists a partition $P$ of $[a, b]$ such that the sum of the lengths of the intervals where the oscillation exceeds $\sigma$ is less than $\epsilon$.
 
-EOP
+QED
 
 #### Proposition 2.4
 

pages/Math4121/Math4121_L15.md

@@ -36,7 +36,7 @@ So $S\setminus \bigcup_{i=1}^{n} I_i$ contains only finitely many points, say $N
 
 So $c_e(S)\leq \ell(C)\leq c_e(S')+\epsilon$.
 
-EOP
+QED
 
 #### Corollary: sef of first species
 

pages/Math4121/Math4121_L16.md

@@ -32,7 +32,7 @@ So $d\in S$, this contradicts the definition of $\beta$ as the supremum of $S$.
 
 So $\beta \geq b$.
 
-EOP
+QED
 
 ### Reviewing sections for Math 4111
 
@@ -91,4 +91,4 @@ Setting $r=\sup a_n$ (by the least upper bound property of real numbers), $r\in
 
 This contradicts the assumption that $a_n,b_n$ as the first element in the list.
 
-EOP
+QED

pages/Math4121/Math4121_L17.md

@@ -38,7 +38,7 @@ If $b\in T$, then by definition of $T$, $b \notin \psi(b)$, but $\psi(b) = T$, w
 
 If $b \notin T$, then $b \in \psi(b)$, which is also a contradiction since $b\in T$. Therefore, $2^S$ cannot have the same cardinality as $S$.
 
-EOP
+QED
 
 ### Back to Hankel's Conjecture
 

pages/Math4121/Math4121_L2.md

@@ -39,7 +39,7 @@ $$
 
 So $h'(x)=\frac{h(t)-h(x)}{t-x}=(f'(x)+u(t))(g'(y)+v(s))$. Since $u(t)\to 0$ and $v(s)\to 0$ as $t\to x$ and $s\to y$, we have $h'(x)=g'(y)f'(x)$.
 
-EOP
+QED
 
 #### Example 5.6
 
@@ -135,4 +135,4 @@ If $x<t<x+\delta$, then $f(x)\geq f(t)$ so $\frac{f(t)-f(x)}{t-x}\geq 0$.
 
 So $\lim_{t\to x}\frac{f(t)-f(x)}{t-x}=0$.
 
-EOP
+QED

pages/Math4121/Math4121_L3.md

@@ -39,7 +39,7 @@ Consider three cases:
 
 In all cases, $h$ has a local minimum or maximum in $(a,b)$.
 
-EOP
+QED
 
 #### Theorem 5.10 Mean Value Theorem
 

pages/Math4121/Math4121_L4.md

@@ -32,7 +32,7 @@ Hence, $g$ attains its infimum on $[a,b]$ at some $x\in (a,b)$. Then this $x$ is
 
 So $g'(x)=0$ and $f'(x)=\lambda$.
 
-EOP
+QED
 
 ### L'Hôpital's Rule
 
@@ -136,4 +136,4 @@ $$
 
 $\forall x\in (a,c_2)$, $\frac{f(x)}{g(x)}<q$.
 
-EOP
+QED

pages/Math4121/Math4121_L5.md

@@ -34,7 +34,7 @@ Set $F(x)=-f(x)$. and $q=-A+\epsilon>-A$. Apply main step, $\exists c_2\in (a,b)
 
 We take $c=\min(c_1,c_2)$. Then $\forall x\in (a,c)$, $\frac{f(x)}{g(x)}<q$.
 
-EOP
+QED
 
 ### Higher Order Derivatives
 
@@ -114,6 +114,6 @@ By Mean Value Theorem, $\exists x_n\in (\alpha,x_{n-1})$ such that $g^{(n)}(x_n)
 
 Since $g^{(n)}(\alpha)=0$ for $k=0,1,2,\cdots,n-1$, we can find $x_n\in (\alpha,x_{n-1})$ such that $g^{(n)}(x_n)=0$.
 
-EOP
+QED
 
 ## Chapter 6: Riemann-Stieltjes Integral

pages/Math4121/Math4121_L6.md

@@ -95,7 +95,7 @@ $$
 
 Same for $U(P_k,f,\alpha)\geq U(P_{k-1},f,\alpha)$.
 
-EOP
+QED
 
 #### Theorem 6.5
 
@@ -119,4 +119,4 @@ $$
 \underline{\int_a^b}f(x)d\alpha\leq \sup_{P_1}L(P_1,f,\alpha)\leq \inf_{P_2}U(P_2,f,\alpha)=\overline{\int_a^b}f(x)d\alpha
 $$
 
-EOP
+QED

pages/Math4121/Math4121_L7.md

@@ -50,7 +50,7 @@ $$
 
 So $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
 
-EOP
+QED
 
 #### Theorem 6.8
 
@@ -93,4 +93,4 @@ $$
 
 So, $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
 
-EOP
+QED

pages/Math4121/Math4121_L8.md

@@ -40,7 +40,7 @@ $$
 
 Therefore, $U(P,f,\alpha) - L(P,f,\alpha)<U(P,\alpha,f) - L(P,\alpha,f) < \epsilon$, so $f\in \mathscr{R}(\alpha)$ on $[a, b]$.
 
-EOP
+QED
 
 #### Theorem 6.10
 
@@ -76,7 +76,7 @@ Since $\epsilon$ is arbitrary, we have $U(P,f,\alpha) - L(P,f,\alpha) < \epsilon
 
 Therefore, $f\in \mathscr{R}(\alpha)$ on $[a, b]$.
 
-EOP
+QED
 
 #### Theorem 6.11
 

pages/Math4121/Math4121_L9.md

@@ -53,7 +53,7 @@ $$
 
 Since $\epsilon$ is arbitrary, $h\in \mathscr{R}(\alpha)$ on $[a, b]$.
 
-EOP
+QED
 
 ### Properties of Integrable Functions
 
@@ -108,4 +108,4 @@ So $U(P,h,\alpha)\leq U(P,f,\alpha)+U(P,g,\alpha)\leq \int_a^b f d\alpha + \int_
 
 Since $\epsilon$ is arbitrary, $\int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha$.
 
-EOP
+QED

pages/Math416/Math416_L1.md

@@ -109,8 +109,32 @@ $$
 
 (Define $\text{cis}(\theta)=\cos\theta+i\sin\theta$)
 
+#### Theorem 1.6 Parallelogram Equality
+
+The sum of the squares of the lengths of the diagonals of a parallelogram equals the sum of the squares of the lengths of the sides.
+
+Proof:
+
+Let $z_1,z_2$ be two complex numbers representing the two sides of the parallelogram, then the sum of the squares of the lengths of the diagonals of the parallelogram is $|z_1-z_2|^2+|z_1+z_2|^2$, and the sum of the squares of the lengths of the sides is $2|z_1|^2+2|z_2|^2$.
+
+$$
+\begin{aligned}
+|z_1-z_2|^2+|z_1+z_2|^2 &= (x_1-x_2)^2+(y_1-y_2)^2+(x_1+x_2)^2+(y_1+y_2)^2 \\
+&= 2x_1^2+2x_2^2+2y_1^2+2y_2^2 \\
+&= 2(|z_1|^2+|z_2|^2)
+\end{aligned}
+$$
+
+QED
+
+#### Definition 1.9
+
+The argument of a complex number $z$ is defined as the angle $\theta$ between the positive real axis and the ray from the origin through $z$.
+
 ### De Moivre's Formula
 
+#### Theorem 1.10 De Moivre's Formula
+
 Let $z=r\text{cis}(\theta)$, then
 
 $\forall n\in \mathbb{Z}$:

pages/Math416/Math416_L10.md

@@ -34,7 +34,7 @@ $$
 \left|\sum_{n=1}^{K}a_n-\sum_{n=1}^{L}a_{\sigma(n)}\right|<2\epsilon
 $$
 
-EOP
+QED
 
 ## Chapter 4 Complex Integration
 
@@ -72,7 +72,7 @@ $$
 
 Assume $\phi$ is continuous on $[a,b]$, the equality means $\lambda(t)\phi(t)$ is real and positive everywhere on $[a,b]$, which means $\arg \phi(t)$ is constant.
 
-EOP
+QED
 
 #### Definition 6.4 Arc Length
 
@@ -82,12 +82,12 @@ $$
 \Gamma=\int_a^b |\gamma'(t)| dt=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt
 $$
 
-N.B. If $\int_{\Gamma} f(\zeta) d\zeta$ depends on orientation of $\Gamma$, but not the parametrization.
+N.B. If $\int_{\Gamma} f(z) dz$ depends on orientation of $\Gamma$, but not the parametrization.
 
 We define
 
 $$
-\int_{\Gamma} f(\zeta) d\zeta=\int_{\Gamma} f(\gamma(t))\gamma'(t) dt
+\int_{\Gamma} f(z) dz=\int_{\Gamma} f(\gamma(t))\gamma'(t) dt
 $$
 
 Example:
@@ -95,7 +95,7 @@ Example:
 Suppose $\Gamma$ is the circle centered at $z_0$ with radius $R$
 
 $$
-\int_{\Gamma} \frac{1}{\zeta-z_0} d\zeta
+\int_{\Gamma} \frac{1}{z-z_0} dz
 $$
 
 Parameterize the unit circle:
@@ -106,7 +106,7 @@ $$
 $$
 
 $$
-f(\zeta)=\frac{1}{\zeta-z_0}
+f(z)=\frac{1}{z-z_0}
 $$
 
 $$
@@ -114,7 +114,7 @@ f(\gamma(t))=\frac{1}{(z_0+Re^{it})-z_0}
 $$
 
 $$
-\int_{\Gamma} f(\zeta) d\zeta=\int_0^{2\pi} f(\gamma(t))\gamma'(t) dt=\int_0^{2\pi} \frac{1}{Re^{-it}}iRe^{it} dt=2\pi i
+\int_{\Gamma} f(z) dz=\int_0^{2\pi} f(\gamma(t))\gamma'(t) dt=\int_0^{2\pi} \frac{1}{Re^{-it}}iRe^{it} dt=2\pi i
 $$
 
 #### Theorem 6.11 (Uniform Convergence)
@@ -142,7 +142,7 @@ $$
 \end{aligned}
 $$
 
-EOP
+QED
 
 #### Theorem 6.6 (Integral of derivative)
 
@@ -157,20 +157,20 @@ $$
 \end{aligned}
 $$
 
-EOP
+QED
 
 Example:
 
 Let $R$ be a rectangle $\{-a,a,ai+b,ai-b\}$, $\Gamma$ is the boundary of $R$ with positive orientation.
 
-Let $\int_{R} e^{-\zeta^2}d\zeta$.
+Let $\int_{R} e^{-z^2}dz$.
 
-Is $e^{-\zeta^2}=\frac{d}{d\zeta}f(\zeta)$?
+Is $e^{-z^2}=\frac{d}{dz}f(z)$?
 
 Yes, since
 
 $$
-e^{\zeta^2}=1-\frac{\zeta^2}{1!}+\frac{\zeta^4}{2!}-\frac{\zeta^6}{3!}+\cdots=\frac{d}{d\zeta}\left(\frac{\zeta}{1!}-\frac{1}{3}\frac{\zeta^3}{2!}+\frac{1}{5}\frac{\zeta^5}{3!}-\cdots\right)
+e^{z^2}=1-\frac{z^2}{1!}+\frac{z^4}{2!}-\frac{z^6}{3!}+\cdots=\frac{d}{dz}\left(\frac{z}{1!}-\frac{1}{3}\frac{z^3}{2!}+\frac{1}{5}\frac{z^5}{3!}-\cdots\right)
 $$
 
 This is polynomial, therefore holomorphic.
@@ -178,7 +178,7 @@ This is polynomial, therefore holomorphic.
 So
 
 $$
-\int_{R} e^{\zeta^2}d\zeta = 0
+\int_{R} e^{z^2}dz = 0
 $$
 
 with some limit calculation, we can get
@@ -186,5 +186,5 @@ with some limit calculation, we can get
 <!--TODO: Fill the parts-->
 
 $$
-\int_{R} e^{-\zeta^2}d\zeta = 2\pi i
+\int_{R} e^{-z^2}dz = 2\pi i
 $$

pages/Math416/Math416_L11.md

@@ -4,23 +4,23 @@
 
 ### Continue on last example
 
-Last lecture we have:Let $R$ be a rectangular start from the $-a$ to $a$, $a+ib$ to $-a+ib$, $\int_{R} e^{-\zeta^2}d\zeta=0$, however, the integral consists of four parts:
+Last lecture we have:Let $R$ be a rectangular start from the $-a$ to $a$, $a+ib$ to $-a+ib$, $\int_{R} e^{-z^2}dz=0$, however, the integral consists of four parts:
 
 Path 1: $-a\to a$
 
-$\int_{I_1}e^{-\zeta^2}d\zeta=\int_{-a}^{a}e^{-\zeta^2}d\zeta=\int_{-a}^{a}e^{-x^2}dx$
+$\int_{I_1}e^{-z^2}dz=\int_{-a}^{a}e^{-z^2}dz=\int_{-a}^{a}e^{-x^2}dx$
 
 Path 2: $a+ib\to -a+ib$
 
-$\int_{I_2}e^{-\zeta^2}d\zeta=\int_{a+ib}^{-a+ib}e^{-\zeta^2}d\zeta=\int_{0}^{b}e^{-(a+iy)^2}dy$
+$\int_{I_2}e^{-z^2}dz=\int_{a+ib}^{-a+ib}e^{-z^2}dz=\int_{0}^{b}e^{-(a+iy)^2}dy$
 
 Path 3: $-a+ib\to -a-ib$
 
-$-\int_{I_3}e^{-\zeta^2}d\zeta=-\int_{-a+ib}^{-a-ib}e^{-\zeta^2}d\zeta=-\int_{a}^{-a}e^{-(x-ib)^2}dx$
+$-\int_{I_3}e^{-z^2}dz=-\int_{-a+ib}^{-a-ib}e^{-z^2}dz=-\int_{a}^{-a}e^{-(x-ib)^2}dx$
 
 Path 4: $-a-ib\to a-ib$
 
-$-\int_{I_4}e^{-\zeta^2}d\zeta=-\int_{-a-ib}^{a-ib}e^{-\zeta^2}d\zeta=-\int_{b}^{0}e^{-(-a+iy)^2}dy$
+$-\int_{I_4}e^{-z^2}dz=-\int_{-a-ib}^{a-ib}e^{-z^2}dz=-\int_{b}^{0}e^{-(-a+iy)^2}dy$
 
 > #### The reverse of a curve 6.9
 >
@@ -37,12 +37,12 @@ If we keep $b$ fixed, and let $a\to\infty$, then
 >
 > Then
 >
-> $$\left|\int_{\gamma}f(\zeta)d\zeta\right|\leq L(\gamma)M$$
+> $$\left|\int_{\gamma}f(z)dz\right|\leq L(\gamma)M$$
 
 _Continue on previous example, we have:_
 
 $$
-\left|\int_{\gamma}f(\zeta)d\zeta\right|\leq L(\gamma)\max_{\zeta\in\gamma}|f(\zeta)|\to 0
+\left|\int_{\gamma}f(z)dz\right|\leq L(\gamma)\max_{z\in\gamma}|f(z)|\to 0
 $$
 
 Since,
@@ -89,7 +89,7 @@ $$
 J=\sqrt{\pi}
 $$
 
-EOP
+QED
 
 ## Chapter 7 Cauchy's theorem
 
@@ -109,7 +109,7 @@ Cauchy's theorem is true if $\gamma$ is a triangle.
 
 Proof:
 
-We plan to keep shrinking the triangle until $f(\zeta+h)=f(\zeta)+hf'(\zeta)+\epsilon(h)$ where $\epsilon(h)$ is a function of $h$ that goes to $0$ as $h\to 0$.
+We plan to keep shrinking the triangle until $f(z+h)=f(z)+hf'(z)+\epsilon(h)$ where $\epsilon(h)$ is a function of $h$ that goes to $0$ as $h\to 0$.
 
 Let's start with a triangle $T$ with vertices $z_1,z_2,z_3$.
 
@@ -127,21 +127,21 @@ Since $L(T_1)=\frac{1}{2}L(T)$, we iterate after $n$ steps, get a triangle $T_n$
 
 Since $K_n=T_n\cup \text{interior}(T_n)$ is compact, we can find $K_n+1\subset K_n$ and $diam(K_n+1)<\frac{1}{2}diam(K_n)$. $diam(K_n)\to 0$ as $n\to\infty$. (Using completeness theorem)
 
-Since $f$ is holomorphic on $u$, $\lim_{\zeta\to z_0}\frac{f(\zeta)-f(z_0)}{\zeta-z_0}=f'(z_0)$ exists.
+Since $f$ is holomorphic on $u$, $\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}=f'(z_0)$ exists.
 
-So $f(\zeta)=f(z_0)+f'(z_0)(\zeta-z_0)+R(\zeta)$, we have
+So $f(z)=f(z_0)+f'(z_0)(z-z_0)+R(z)$, we have
 
 $$
-\int_{T_n}f(\zeta)d\zeta=\int_{T_n}f(z_0)d\zeta+\int_{T_n}f'(z_0)(\zeta-z_0)d\zeta+\int_{T_n}R(\zeta)d\zeta
+\int_{T_n}f(z)dz=\int_{T_n}f(z_0)dz+\int_{T_n}f'(z_0)(z-z_0)dz+\int_{T_n}R(z)dz
 $$
 
-since $f(z_0)d\zeta+\int_{T_n}f'(z_0)(\zeta-z_0)$ is in form of Cauchy integral formula, we have
+since $f(z_0)dz+\int_{T_n}f'(z_0)(z-z_0)$ is in form of Cauchy integral formula, we have
 
 $$
-\int_{T_n}f(z_0)d\zeta+\int_{T_n}f'(z_0)(\zeta-z_0)d\zeta=0
+\int_{T_n}f(z_0)dz+\int_{T_n}f'(z_0)(z-z_0)dz=0
 $$
 
-Let $e_n=\max\{\frac{R(\zeta)}{\zeta-z_0}:z_0\in T_n\}$
+Let $e_n=\max\{\frac{R(z)}{z-z_0}:z_0\in T_n\}$
 
 Since $diam(K_n)\to 0$ as $n\to\infty$, we have $e_n\to 0$ as $n\to\infty$.
 
@@ -149,9 +149,9 @@ So
 
 $$
 \begin{aligned}
-|I|&\leq 4^n\left|\int_{T_n}f(\zeta)d\zeta\right|\\
-&\leq 4^n\left|\int_{T_n}R_n(\zeta)d\zeta\right|\\
-&\leq 4^n\cdot L(T_n)\cdot \max_{\zeta\in T_n}|R_n(\zeta)|\\
+|I|&\leq 4^n\left|\int_{T_n}f(z)dz\right|\\
+&\leq 4^n\left|\int_{T_n}R_n(z)dz\right|\\
+&\leq 4^n\cdot L(T_n)\cdot \max_{z\in T_n}|R_n(z)|\\
 &\leq 4^n\cdot \frac{L(T_0)}{2^n}\cdot e_n L(T_n)\\
 &\leq 4^n\cdot \frac{L(T_0)}{2^n}\cdot e_n\cdot \frac{L(T_0)}{2^n}\\
 &\leq e_n\cdot L(T_0)^2
@@ -163,7 +163,7 @@ Since $e_n\to 0$ as $n\to\infty$, we have $I\to 0$ as $n\to\infty$.
 So
 
 $$
-\int_{T_n}f(\zeta)d\zeta\to 0
+\int_{T_n}f(z)dz\to 0
 $$
 
-EOP
+QED

pages/Math416/Math416_L12.md

@@ -7,114 +7,114 @@
 Let $T$ be a triangle in $\mathbb{C}$ and $f$ be holomorphic on $T$. Then
 
 $$
-\int_T f(\zeta) d\zeta = 0
+\int_T f(z) dz = 0
 $$
 
 ### Cauchy's Theorem for Convex Sets
 
-Let's start with a simple case: $f(\zeta)=1$.
+Let's start with a simple case: $f(z)=1$.
 
 For any closed curve $\gamma$ in $U$, we have
 
 $$
-\int_\gamma f(\zeta) d\zeta = \int_a^b f(\gamma(t)) \gamma'(t) dt \approx \sum_{i=1}^n f(\gamma(t_i)) \gamma'(t_i) \Delta t_i
+\int_\gamma f(z) dz = \int_a^b f(\gamma(t)) \gamma'(t) dt \approx \sum_{i=1}^n f(\gamma(t_i)) \gamma'(t_i) \Delta t_i
 $$
 
 #### Definition of a convex set
 
-A set $U$ is convex if for any two points $\zeta_1, \zeta_2 \in U$, the line segment $[\zeta_1, \zeta_2] \subset U$.
+A set $U$ is convex if for any two points $z_1, z_2 \in U$, the line segment $[z_1, z_2] \subset U$.
 
 Let $O(U)$ be the set of all holomorphic functions on $U$.
 
 #### Definition of primitive
 
-Say $f$ has a primitive on $U$. If there exists a holomorphic function $g$ on $U$ such that $g'(\zeta)=f(\zeta)$ for all $\zeta \in U$, then $g$ is called a primitive of $f$ on $U$.
+Say $f$ has a primitive on $U$. If there exists a holomorphic function $g$ on $U$ such that $g'(z)=f(z)$ for all $z \in U$, then $g$ is called a primitive of $f$ on $U$.
 
 #### Cauchy's Theorem for a Convex region
 
 Cauchy's Theorem holds if $f$ has a primitive on a convex region $U$.
 
 $$
-\int_\gamma f(\zeta) d\zeta = \int_\gamma \left[\frac{d}{d\zeta}g(\zeta)\right] d\zeta = g(\zeta_1)-g(\zeta_2)
+\int_\gamma f(z) dz = \int_\gamma \left[\frac{d}{dz}g(z)\right] dz = g(z_1)-g(z_2)
 $$
 
-Since the curve is closed, $\zeta_1=\zeta_2$, so $\int_\gamma f(\zeta) d\zeta = 0$.
+Since the curve is closed, $z_1=z_2$, so $\int_\gamma f(z) dz = 0$.
 
 Proof:
 
 It is sufficient to prove that if $U$ is convex, $f$ is holomorphic on $U$, then $f=g'$ for some $g$ holomorphic on $U$.
 
-We pick a point $z_0\in U$ and define $g(\zeta)=\int_{[\zeta_0,\zeta]}f(\xi)d\xi$.
+We pick a point $z_0\in U$ and define $g(z)=\int_{[z_0,z]}f(\xi)d\xi$.
 
 We claim $g\in O(U)$ and $g'=f$.
 
-Let $\zeta_1$ close to $\zeta$, since $f$ is holomorphic on $U$, using the Goursat's theorem, we can find a triangle $T$ with $\xi\in T$ and $\zeta\in T$ and $T\subset U$.
+Let $z_1$ close to $z$, since $f$ is holomorphic on $U$, using the Goursat's theorem, we can find a triangle $T$ with $\xi\in T$ and $z\in T$ and $T\subset U$.
 
 $$
 \begin{aligned}
-0&=\int_{\zeta_0}^{\zeta}f(\xi)d\xi+\int_{\zeta}^{\zeta_1}f(\xi)d\xi\\
-&=g(\zeta)-g(\zeta_1)+\int_{\zeta}^{\zeta_1}f(\xi)d\xi+\int_{\zeta_1}^{\zeta_0}f(\xi)d\xi\\
-\frac{g(\zeta)-g(\zeta_1)}{\zeta-\zeta_1}&=-\frac{1}{\zeta-\zeta_1}\left(\int_{\zeta}^{\zeta_1}f(\xi)d\xi\right)\\
-\frac{g(\zeta_1)-g(\zeta_0)}{\zeta_1-\zeta_0}-f(\zeta_1)&=-\frac{1}{\zeta_1-\zeta_0}\left(\int_{\zeta}^{\zeta_1}f(\xi)d\xi\right)-f(\zeta_1)\\
-&=-\frac{1}{\zeta_1-\zeta_0}\left(\int_{\zeta}^{\zeta_1}f(\xi)-f(\zeta_1)d\xi\right)\\
+0&=\int_{z_0}^{z}f(\xi)d\xi+\int_{z}^{z_1}f(\xi)d\xi\\
+&=g(z)-g(z_1)+\int_{z}^{z_1}f(\xi)d\xi+\int_{z_1}^{z_0}f(\xi)d\xi\\
+\frac{g(z)-g(z_1)}{z-z_1}&=-\frac{1}{z-z_1}\left(\int_{z}^{z_1}f(\xi)d\xi\right)\\
+\frac{g(z_1)-g(z_0)}{z_1-z_0}-f(z_1)&=-\frac{1}{z_1-z_0}\left(\int_{z}^{z_1}f(\xi)d\xi\right)-f(z_1)\\
+&=-\frac{1}{z_1-z_0}\left(\int_{z}^{z_1}f(\xi)-f(z_1)d\xi\right)\\
 &=I
 \end{aligned}
 $$
 
-Use the fact that $f$ is holomorphic on $U$, then $f$ is continuous on $U$, so $\lim_{\zeta\to\zeta_1}f(\zeta)=f(\zeta_1)$.
+Use the fact that $f$ is holomorphic on $U$, then $f$ is continuous on $U$, so $\lim_{z\toz_1}f(z)=f(z_1)$.
 
-There exists a $\delta>0$ such that $|\zeta-\zeta_1|<\delta$ implies $|f(\zeta)-f(\zeta_1)|<\epsilon$.
+There exists a $\delta>0$ such that $|z-z_1|<\delta$ implies $|f(z)-f(z_1)|<\epsilon$.
 
 So
 
 $$
-|I|\leq\frac{1}{\zeta_1-\zeta_0}\int_{\zeta}^{\zeta_1}|f(\xi)-f(\zeta_1)|d\xi<\frac{\epsilon}{\zeta_1-\zeta_0}\int_{\zeta}^{\zeta_1}d\xi=\epsilon
+|I|\leq\frac{1}{z_1-z_0}\int_{z}^{z_1}|f(\xi)-f(z_1)|d\xi<\frac{\epsilon}{z_1-z_0}\int_{z}^{z_1}d\xi=\epsilon
 $$
 
-So $I\to 0$ as $\zeta_1\to\zeta$.
+So $I\to 0$ as $z_1\toz$.
 
-Therefore, $g'(\zeta_1)=f(\zeta_1)$ for all $\zeta_1\in U$.
+Therefore, $g'(z_1)=f(z_1)$ for all $z_1\in U$.
 
-EOP
+QED
 
 ### Cauchy's Theorem for a disk
 
-Let $U$ be the open set, $f\in O(U)$. Let $C$ be a circle inside $U$ and $\zeta$ be a point inside $C$.
+Let $U$ be the open set, $f\in O(U)$. Let $C$ be a circle inside $U$ and $z$ be a point inside $C$.
 
 Then
 
 $$
-f(\zeta)=\frac{1}{2\pi i}\int_C\frac{f(\xi)d\xi}{\xi-\zeta} d\xi
+f(z)=\frac{1}{2\pi i}\int_C\frac{f(\xi)d\xi}{\xi-z} d\xi
 $$
 
 Proof:
 
-Let $C_\epsilon$ be a circle with center $\zeta$ and radius $\epsilon$ inside $C$.
+Let $C_\epsilon$ be a circle with center $z$ and radius $\epsilon$ inside $C$.
 
 Claim:
 
 $$
-\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=\int_{C}\frac{f(\xi)d\xi}{\xi-\zeta}
+\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-z}=\int_{C}\frac{f(\xi)d\xi}{\xi-z}
 $$
 
 We divide the integral into four parts:
 
 ![Integral on a disk](https://notenextra.trance-0.com/Math416/Cauchy_disk.png)
 
-Notice that $\frac{f(\xi)}{\xi-\zeta}$ is holomorphic whenever $f(\xi)\in U$ and $\xi\in \mathbb{C}\setminus\{\zeta\}$.
+Notice that $\frac{f(\xi)}{\xi-z}$ is holomorphic whenever $f(\xi)\in U$ and $\xi\in \mathbb{C}\setminus\{z\}$.
 
 So we can apply Cauchy's theorem to the integral on the inside square.
 
 $$
-\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=0
+\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-z}=0
 $$
 
-Since $\frac{1}{2\pi i}\int_{C_\epsilon}\frac{1}{\xi-\zeta}d\xi=1$, $\sigma=\epsilon e^{it}+\zeta_0$ and $\sigma'=\epsilon e^{it}$, we have
+Since $\frac{1}{2\pi i}\int_{C_\epsilon}\frac{1}{\xi-z}d\xi=1$, $\sigma=\epsilon e^{it}+z_0$ and $\sigma'=\epsilon e^{it}$, we have
 
 /* TRACK LOST*/
 
 $$
-\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=\int_0^{2\pi}\frac{f(\sigma)d\sigma}{\sigma-\zeta}=2\pi i f(\zeta)
+\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-z}=\int_0^{2\pi}\frac{f(\sigma)d\sigma}{\sigma-z}=2\pi i f(z)
 $$
 
-EOP
+QED

pages/Math416/Math416_L13.md

@@ -0,0 +1,182 @@
+# Math416 Lecture 13
+
+## Review on Cauchy's Theorem
+
+Cauchy's Theorem states that if a function is holomorphic (complex differentiable) on a simply connected domain, then the integral of that function over any closed contour within that domain is zero.
+
+Last lecture we proved the case for convex regions.
+
+### Cauchy's Formula for a Circle
+
+Let $C$ be a counterclockwise oriented circle, and let $f$ be a holomorphic
+
+function defined in an open set containing $C$ and its interior. Then,
+
+$$
+f(z_0)=\frac{1}{2\pi i}\int_C\frac{f(z)}{z-z_0}dz
+$$
+
+for all points $z$ in the interior of $C$.
+
+## New materials
+
+### Mean value property
+
+#### Theorem 7.6: Mean value property
+
+Special case: Suppose $f$ is holomorphic on some $\mathbb{D}(z_0,R)\subset \mathbb{C}$, by cauchy's formula,
+
+$$
+f(z_0)=\frac{1}{2\pi i}\int_{C_r}\frac{f(z)}{z-z_0}dz
+$$
+
+Parameterizing $C_r$, we get $\gamma(t)=z_0+re^{it}$, $0\leq t\leq 2\pi$
+
+$$
+\int f(z)dz=\int f(\gamma) \gamma'(t) d t
+$$
+
+So,
+
+$$
+f(z_0)=\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(z_0+re^{it})}{re^{it}} ire^{it} dt=\frac{1}{2\pi}\int_{0}^{2\pi} f(z_0+re^{it}) dt
+$$
+
+This concludes the mean value property for the holomorphic function
+
+If $f$ is holomorphic, $f(z_0)$ is the mean value of $f$ on any circle centered at $z_0$
+
+#### Area representation of mean value property
+
+Area of $f$ on $\mathbb{D}(z_0,r)$
+
+$$
+\frac{1}{\pi r^2}\int_{0}^{2\pi}\int_0^r f(z_0+re^{it})
+$$
+
+/*Track lost*/
+
+### Cauchy Integral
+
+#### Definition 7.7
+
+Let $\gamma:[a,b]\to \mathbb{C}$ be piecewise $\mathbb{C}^1$, let $\phi$ be condition on $\gamma$. Then the Cauchy interval of $\phi$ along $\gamma$ is 
+
+$$
+F(z)=\int_{\gamma}\frac{\phi(\zeta)}{\zeta-z}d \zeta
+$$
+
+#### Theorem
+
+Suppose $F(z)=\int_{\gamma}\frac{\phi(z)}{\zeta-z}d z$. Then $F$ has a local power series representation at all points $z_0$ not in $\gamma$.
+
+Proof:
+
+Let $R=B(z_0,\gamma)>0$, let $z\in \mathbb{D}(z_0,R)$
+
+So
+
+$$
+\frac{1}{\zeta-z}=\frac{1}{(\zeta-z_0)-(z-z_0)}=\frac{1}{1-z_0}\frac{1}{1-\frac{z-z_0}{\zeta-z_0}}
+$$
+
+Since $|z-z_0|<R$ and $|\zeta-z_0|>R$, $|\frac{z-z_0}{\zeta-z_0}|<1$.
+
+Converting it to geometric series
+
+$$
+\frac{1}{1-z_0}\frac{1}{1-\frac{z-z_0}{\zeta-z_0}}=\sum_{n=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^n
+$$
+
+So,
+
+$$
+\begin{aligned}
+F(z)&=\int_\gamma \frac{\phi(\zeta)}{\zeta - z} d\zeta\\
+&=\int_\gamma \frac{\phi(\zeta)}{z-z_0} \sum_{n=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^n dz\\
+&=\sum_{n=0}^\infty (z-z_0)^n \int_\gamma \frac{\phi(\zeta)}{(\zeta-z_0)^{n+1}} 
+\end{aligned}
+$$
+
+Which gives us an power series representation if we set $a_n=\int_\gamma \frac{\phi(\zeta)}{(\zeta-z_0)^{n+1}}$
+
+QED
+
+#### Corollary 7.7
+
+Suppose $F(z)=\int_\gamma \frac{\phi(\zeta)}{\zeta-z_0} dz$,
+
+Then,
+
+$$
+f^{(n)}(z)=n!\int_\gamma \frac{\phi(z)}{(\zeta-z_0)^{n+1}} d\zeta
+$$
+
+where $z\in \mathbb{C}\setminus \gamma$.
+
+Combine with Cauchy integral formula:
+
+If $f$ is in $O(\Omega)$, then $\forall z\in \mathbb{D}(z_0,r)$.
+
+$$
+f(z)=\frac{1}{2\pi i}\int_{C_r}\frac{f(\zeta)}{\zeta-z} d\zeta
+$$
+
+We have proved that If $f\in O(\Omega)$, then $f$ is locally given  by a convergent power series
+
+power series has radius of convergence at $z_0$ that is $\geq$ dist($z_0$,boundary $\Omega$)
+
+### Liouville's Theorem
+
+#### Definition 7.11
+
+A function that is holomorphic in all of $\mathbb{C}$ is called an entire function.
+
+#### Theorem 7.11 Liouville's Theorem
+
+Any bounded entire function is constant.
+
+> Basic Estimate of integral
+>
+> $$\left|\int_\gamma f(z) dz\right|\leq L(\gamma) \max\left|f(z)\right|$$
+
+Since,
+
+$$
+f'(z)=\frac{1}{2\pi i} \int_{C_r} \frac{f(z)}{(\zeta-z)^2} dz
+$$
+
+So the modulus of the integral is bounded by
+
+$$
+\frac{1}{2\pi} |M|\cdot \frac{1}{R^2}=2\pi R\cdot M \frac{1}{R^2}=\frac{M}{R}
+$$
+
+### Fundamental Theorem of Algebra
+
+#### Theorem 7.12 Fundamental Theorem of ALgebra
+
+Every nonconstant polynomial with complex coefficients can be factored over
+$\mathbb{C}$ into linear factors.
+
+#### Corollary
+
+For every polynomial with complex coefficients.
+
+$$
+p(z)=c\prod_{j=i}^n(z-z_0)^{t_j}
+$$
+
+where the degree of polynomial is $\sum_{j=0}^n t_j$
+
+Proof:
+
+Let $p(z)=a_0+a_1z+\cdots+a_nz^n$, where $a_n\neq 0$ and $n\geq 1$.
+
+So
+
+$$
+|p(z)|=|a_nz_n|\left[\left|1+\frac{a_{n-1}}{a_nz}+\cdots+\frac{a_0}{a_nz^n}\right|\right]
+$$
+
+If $|z|\geq R$, $\left|1+\frac{a_{n-1}}{a_nz}+\cdots+\frac{a_0}{a_nz^n}\right|<\frac{1}{2}$

pages/Math416/Math416_L2.md

@@ -24,6 +24,7 @@ $$
 \forall n\in \mathbb{Z}, z^n=r^n\text{cis}(n\theta)
 $$
 
+
 ## New Fancy stuff
 
 Claim:
@@ -54,7 +55,7 @@ When $k=1$, we get $\text{cis}\left(\frac{2\pi}{3}\right)=-\frac{1}{2}+i\frac{\s
 
 When $k=2$, we get $\text{cis}\left(\frac{4\pi}{3}\right)=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$
 
-#### Strange example
+Strange example
 
 Let $p(x)=a_3x^3+a_2x^2+a_1x+a_0$ be a polynomial with real coefficients.
 
@@ -197,18 +198,18 @@ So all the point on the north pole is mapped to outside of the unit circle in $\
 
 all the point on the south pole is mapped to inside of the unit circle in $\mathbb{R}^2$.
 
-The line through $(0,0,1)$ and $(\xi,\eta,\zeta)$ intersects the unit sphere at $(x,y,0)$
+The line through $(0,0,1)$ and $(\xi,\eta,z)$ intersects the unit sphere at $(x,y,0)$
 
-Line $(tx,ty,1-t)$ intersects $\zeta^2$ at $t^2x^2+t^2y^2+(1-t)^2=1$
+Line $(tx,ty,1-t)$ intersects $z^2$ at $t^2x^2+t^2y^2+(1-t)^2=1$
 
 So $t=\frac{2}{1+x^2+y^2}$
 
 $$
-\zeta=x+iy\mapsto \frac{1}{1+|\zeta|^2}(2Re(\zeta),2Im(\zeta),|\zeta|^2-1)
+z=x+iy\mapsto \frac{1}{1+|z|^2}(2Re(z),2Im(z),|z|^2-1)
 $$
 
 $$
-(\xi,\eta,\zeta)\mapsto \frac{\xi+i\eta}{1-\zeta}
+(\xi,\eta,z)\mapsto \frac{\xi+i\eta}{1-z}
 $$
 
 This is a homeomorphism. $\mathbb{C}\setminus\{\infty\}\simeq S^2$
@@ -220,7 +221,7 @@ Suppose $\Omega$ is an open subset of $\mathbb{C}$.
 A function $f:\Omega\to \mathbb{C}$'s derivative is defined as
 
 $$
-f'(\zeta_0)=\lim_{\zeta\to \zeta_0}\frac{f(\zeta)-f(\zeta_0)}{\zeta-\zeta_0}
+f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}
 $$
 
 $f=u+iv$, $u,v:\Omega\to \mathbb{R}$
@@ -232,11 +233,11 @@ How are $f'$ and derivatives of $u$ and $v$ related?
 Chain rule applies
 
 $$
-\frac{d}{d\zeta}(f(g(\zeta)))=f'(g(\zeta))g'(\zeta)
+\frac{d}{dz}(f(g(z)))=f'(g(z))g'(z)
 $$
 
 Polynomials
 
 $$
-\frac{d}{d\zeta}\zeta^n=n\zeta^{n-1}
+\frac{d}{dz}z^n=nz^{n-1}
 $$

pages/Math416/Math416_L3.md

@@ -4,14 +4,14 @@
 
 ### Differentiability
 
-#### Definition of differentiability in complex variables
+#### Definition 2.1 of differentiability in complex variables
 
-**Suppose $G$ is an open subset of $\mathbb{C}$**.
+**Suppose $G$ is an open subset of $\mathbb{C}$**. (very important, $f'(z_0)$ cannot be define unless $z_0$ belongs to an open set in which $f$ is defined.)
 
-A function $f:G\to \mathbb{C}$ is differentiable at $\zeta_0\in G$ if
+A function $f:G\to \mathbb{C}$ is differentiable at $z_0\in G$ if
 
 $$
-f'(\zeta_0)=\lim_{\zeta\to \zeta_0}\frac{f(\zeta)-f(\zeta_0)}{\zeta-\zeta_0}
+f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}
 $$
 
 exists.
@@ -20,7 +20,7 @@ Or equivalently,
 
 We can also express the $f$ as $f=u+iv$, where $u,v:G\to \mathbb{R}$ are real-valued functions.
 
-Recall that $u:G\to \mathbb{R}$ is differentiable at $\zeta_0\in G$ if and only if there exists a complex number $(x,y)\in \mathbb{C}$ such that a function
+Recall that $u:G\to \mathbb{R}$ is differentiable at $z_0\in G$ if and only if there exists a complex number $(x,y)\in \mathbb{C}$ such that a function
 
 $$
 R(x,y)=u(x,y)-\left(u(x_0,y_0)+\frac{\partial u}{\partial x}(x_0,y_0)(x-x_0)+\frac{\partial u}{\partial y}(x_0,y_0)(y-y_0)\right)
@@ -65,9 +65,9 @@ Then the differentiability of $v$ at $(x_0,y_0)$ guarantees that
 $$
 \lim_{(x,y)\to (x_0,y_0)}\frac{|S(x,y)|}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0.
 $$
-Moreover, considering the definition of the complex derivative of $f=u+iv$, if we approach $\zeta_0=x_0+iy_0$ along different directions we obtain
+Moreover, considering the definition of the complex derivative of $f=u+iv$, if we approach $z_0=x_0+iy_0$ along different directions we obtain
 $$
-f'(\zeta_0)=\frac{\partial u}{\partial x}(x_0,y_0)+i\frac{\partial v}{\partial x}(x_0,y_0)
+f'(z_0)=\frac{\partial u}{\partial x}(x_0,y_0)+i\frac{\partial v}{\partial x}(x_0,y_0)
 =\frac{\partial v}{\partial y}(x_0,y_0)-i\frac{\partial u}{\partial y}(x_0,y_0).
 $$
 Equating the real and imaginary parts of these two expressions forces
@@ -77,7 +77,7 @@ $$
 
 #### Theorem 2.6 (The Cauchy-Riemann equations):
 
-If $f=u+iv$ is complex differentiable at $\zeta_0\in G$, then $u$ and $v$ are real differentiable at $(x_0,y_0)$ and
+If $f=u+iv$ is complex differentiable at $z_0\in G$, then $u$ and $v$ are real differentiable at $(x_0,y_0)$ and
 
 $$
 \frac{\partial u}{\partial x}(x_0,y_0)=\frac{\partial v}{\partial y}(x_0,y_0),\quad \frac{\partial u}{\partial y}(x_0,y_0)=-\frac{\partial v}{\partial x}(x_0,y_0).
@@ -85,35 +85,37 @@ $$
 
 > Some missing details:
 >
-> The Cauchy-Riemann equations are necessary and sufficient for the differentiability of $f$ at $\zeta_0$.
+> The Cauchy-Riemann equations are necessary and sufficient for the differentiability of $f$ at $z_0$.
 >
-> This states that a function $f$ is **complex differentiable** at $\zeta_0$ if and only if $u$ and $v$ are real differentiable at $(x_0,y_0)$ and the Cauchy-Riemann equations hold at $(x_0,y_0)$. That is $f'(\zeta_0)=\frac{\partial u}{\partial x}(x_0,y_0)+i\frac{\partial v}{\partial x}(x_0,y_0)=\frac{\partial v}{\partial y}(x_0,y_0)-i\frac{\partial u}{\partial y}(x_0,y_0)$.
+> This states that a function $f$ is **complex differentiable** at $z_0$ if and only if $u$ and $v$ are real differentiable at $(x_0,y_0)$ and the Cauchy-Riemann equations hold at $(x_0,y_0)$. That is $f'(z_0)=\frac{\partial u}{\partial x}(x_0,y_0)+i\frac{\partial v}{\partial x}(x_0,y_0)=\frac{\partial v}{\partial y}(x_0,y_0)-i\frac{\partial u}{\partial y}(x_0,y_0)$.
 
 And $u$ and $v$ have continuous partial derivatives at $(x_0,y_0)$.
 
 And let $c=\frac{\partial u}{\partial x}(x_0,y_0)$ and $d=\frac{\partial v}{\partial x}(x_0,y_0)$.
 
-**Then $f'(\zeta_0)=c+id$, is holomorphic at $\zeta_0$.**
+**Then $f'(z_0)=c+id$, is holomorphic at $z_0$.**
 
 ### Holomorphic Functions
 
 #### Definition 2.8 (Holomorphic functions)
 
-A function $f:G\to \mathbb{C}$ is holomorphic (or analytic) at $\zeta_0\in G$ if it is complex differentiable at $\zeta_0$.
+A function $f:G\to \mathbb{C}$ is holomorphic (or analytic) at $z_0\in G$ if it is complex differentiable at $z_0$.
+
+> Note that the true definition of analytic function is that can be written as a convergent power series in a neighborhood of each point in its domain. We will prove that these two definitions are equivalent to each other in later sections.
 
 Example:
 
 Suppose $f:G\to \mathbb{C}$ where $f=u+iv$ and $\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}$, $\frac{\partial f}{\partial y}=\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}$.
 
-Define $\frac{\partial}{\partial \zeta}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)$ and $\frac{\partial}{\partial \bar{\zeta}}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$.
+Define $\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)$ and $\frac{\partial}{\partial \bar{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$.
 
-Suppose $f$ is holomorphic at $\bar{\zeta}_0\in G$ (Cauchy-Riemann equations hold at $\bar{\zeta}_0$).
+Suppose $f$ is holomorphic at $\bar{z}_0\in G$ (Cauchy-Riemann equations hold at $\bar{z}_0$).
 
-Then $\frac{\partial f}{\partial \bar{\zeta}}(\bar{\zeta}_0)=0$.
+Then $\frac{\partial f}{\partial \bar{z}}(\bar{z}_0)=0$.
 
-Note that $\forall m\in \mathbb{Z}$, $\zeta^m$ is holomorphic on $\mathbb{C}$.
+Note that $\forall m\in \mathbb{Z}$, $z^m$ is holomorphic on $\mathbb{C}$.
 
-i.e. $\forall a\in \mathbb{C}$, $\lim_{\zeta\to a}\frac{\zeta^m-a^m}{\zeta-a}=\frac{(\zeta-a)(\zeta^{m-1}+\zeta^{m-2}a+\cdots+a^{m-1})}{\zeta-a}=ma^{m-1}$.
+i.e. $\forall a\in \mathbb{C}$, $\lim_{z\to a}\frac{z^m-a^m}{z-a}=\frac{(z-a)(z^{m-1}+z^{m-2}a+\cdots+a^{m-1})}{z-a}=ma^{m-1}$.
 
 So polynomials are holomorphic on $\mathbb{C}$.
 
@@ -132,20 +134,20 @@ $$
 And
 
 $$
-\frac{\partial}{\partial \zeta}f=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)f,\quad \frac{\partial}{\partial \bar{\zeta}}f=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)f.
+\frac{\partial}{\partial z}f=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)f,\quad \frac{\partial}{\partial \bar{z}}f=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)f.
 $$
 
 This definition of partial differential operators on complex functions is consistent with the definition of partial differential operators on real functions.
 
 $$
-\frac{\partial}{\partial x}f=\frac{\partial}{\partial \zeta}f+\frac{\partial}{\partial \bar{\zeta}}f,\quad \frac{\partial}{\partial y}f=i\left(\frac{\partial}{\partial \zeta}f-\frac{\partial}{\partial \bar{\zeta}}f\right).
+\frac{\partial}{\partial x}f=\frac{\partial}{\partial z}f+\frac{\partial}{\partial \bar{z}}f,\quad \frac{\partial}{\partial y}f=i\left(\frac{\partial}{\partial z}f-\frac{\partial}{\partial \bar{z}}f\right).
 $$
 
 ### Curves in $\mathbb{C}$
 
 #### Definition 2.11 (Curves in $\mathbb{C}$)
 
-A curve $\gamma$ in $G\subset \mathbb{C}$ is a continuous map of an interval $I$ into $G$. We say $\gamma$ is differentiable if $\forall t_0\in I$, $\gamma'(t_0)=\lim_{t\to t_0}\frac{\gamma(t)-\gamma(t_0)}{t-t_0}$ exists.
+A curve $\gamma$ in $G\subset \mathbb{C}$ is a continuous map of an interval $I\in \mathbb{R}$ into $G$. We say $\gamma$ is differentiable if $\forall t_0\in I$, $\gamma'(t_0)=\lim_{t\to t_0}\frac{\gamma(t)-\gamma(t_0)}{t-t_0}$ exists.
 
 If $\gamma'(t_0)$ is a point in $\mathbb{C}$, then $\gamma'(t_0)$ is called the tangent vector to $\gamma$ at $t_0$.
 
@@ -155,19 +157,19 @@ A curve $\gamma$ is regular if $\gamma'(t)\neq 0$ for all $t\in I$.
 
 #### Definition of angle between two curves
 
-Let $\gamma_1,\gamma_2$ be two curves in $G\subset \mathbb{C}$ with $\gamma_1(t_0)=\gamma_2(t_0)=\zeta_0$ for some $t_0\in I_1\cap I_2$.
+Let $\gamma_1,\gamma_2$ be two curves in $G\subset \mathbb{C}$ with $\gamma_1(t_0)=\gamma_2(t_0)=z_0$ for some $t_0\in I_1\cap I_2$.
 
-The angle between $\gamma_1$ and $\gamma_2$ at $\zeta_0$ is the angle between the vectors $\gamma_1'(t_0)$ and $\gamma_2'(t_0)$. Denote as $\arg(\gamma_2'(t_0))-\arg(\gamma_1'(t_0))=\arg(\gamma_2'(t_0)\gamma_1'(t_0))$.
+The angle between $\gamma_1$ and $\gamma_2$ at $z_0$ is the angle between the vectors $\gamma_1'(t_0)$ and $\gamma_2'(t_0)$. Denote as $\arg(\gamma_2'(t_0))-\arg(\gamma_1'(t_0))=\arg(\gamma_2'(t_0)\gamma_1'(t_0))$.
 
-#### Theorem of conformality
+#### Theorem 2.12 of conformality
 
-Suppose $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma_1,\gamma_2$ are regular curves in $G$ with $\gamma_1(t_0)=\gamma_2(t_0)=\zeta_0$ for some $t_0\in I_1\cap I_2$.
+Suppose $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma_1,\gamma_2$ are regular curves in $G$ with $\gamma_1(t_0)=\gamma_2(t_0)=z_0$ for some $t_0\in I_1\cap I_2$.
 
-If $f'(\zeta_0)\neq 0$, then the angle between $\gamma_1$ and $\gamma_2$ at $\zeta_0$ is the same as the angle between the vectors $f'(\zeta_0)\gamma_1'(t_0)$ and $f'(\zeta_0)\gamma_2'(t_0)$.
+If $f'(z_0)\neq 0$, then the angle between $\gamma_1$ and $\gamma_2$ at $z_0$ is the same as the angle between the vectors $f'(z_0)\gamma_1'(t_0)$ and $f'(z_0)\gamma_2'(t_0)$.
 
 #### Lemma of function of a curve and angle
 
-If $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma$ is differentiable curve in $G$ with $\gamma(t_0)=\zeta_0$ for some $t_0\in I$.
+If $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma$ is differentiable curve in $G$ with $\gamma(t_0)=z_0$ for some $t_0\in I$.
 
 Then,
 
@@ -175,7 +177,7 @@ $$
 (f\circ \gamma)'(t_0)=f'(\gamma(t_0))\gamma'(t_0).
 $$
 
-If Lemma of function of a curve and angle holds, then the angle between $f\circ \gamma_1$ and $f\circ \gamma_2$ at $\zeta_0$ is
+If Lemma of function of a curve and angle holds, then the angle between $f\circ \gamma_1$ and $f\circ \gamma_2$ at $z_0$ is
 
 $$
 \begin{aligned}

pages/Math416/Math416_L4.md

@@ -14,9 +14,9 @@ $$
 
 ### Angle between two curves
 
-Let $\gamma_1,\gamma_2$ be two curves in $G\subset \mathbb{C}$ with $\gamma_1(t_0)=\gamma_2(t_0)=\zeta_0$ for some $t_0\in I_1\cap I_2$.
+Let $\gamma_1,\gamma_2$ be two curves in $G\subset \mathbb{C}$ with $\gamma_1(t_0)=\gamma_2(t_0)=z_0$ for some $t_0\in I_1\cap I_2$.
 
-The angle between $\gamma_1$ and $\gamma_2$ at $\zeta_0$ is the angle between the vectors $\gamma_1'(t_0)$ and $\gamma_2'(t_0)$. Denote as $\arg(\gamma_2'(t_0))-\arg(\gamma_1'(t_0))=\arg(\gamma_2'(t_0)\gamma_1'(t_0))$.
+The angle between $\gamma_1$ and $\gamma_2$ at $z_0$ is the angle between the vectors $\gamma_1'(t_0)$ and $\gamma_2'(t_0)$. Denote as $\arg(\gamma_2'(t_0))-\arg(\gamma_1'(t_0))=\arg(\gamma_2'(t_0)\gamma_1'(t_0))$.
 
 ### Cauchy-Riemann equations
 
@@ -28,13 +28,13 @@ $$
 
 ### Theorem of conformality
 
-Suppose $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma_1,\gamma_2$ are regular curves in $G$ with $\gamma_1(t_0)=\gamma_2(t_0)=\zeta_0$ for some $t_0\in I_1\cap I_2$.
+Suppose $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma_1,\gamma_2$ are regular curves in $G$ with $\gamma_1(t_0)=\gamma_2(t_0)=z_0$ for some $t_0\in I_1\cap I_2$.
 
-If $f'(\zeta_0)\neq 0$, then the angle between $\gamma_1$ and $\gamma_2$ at $\zeta_0$ is the same as the angle between the vectors $f'(\zeta_0)\gamma_1'(t_0)$ and $f'(\zeta_0)\gamma_2'(t_0)$.
+If $f'(z_0)\neq 0$, then the angle between $\gamma_1$ and $\gamma_2$ at $z_0$ is the same as the angle between the vectors $f'(z_0)\gamma_1'(t_0)$ and $f'(z_0)\gamma_2'(t_0)$.
 
 ### Lemma of function of a curve and angle
 
-If $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma$ is differentiable curve in $G$ with $\gamma(t_0)=\zeta_0$ for some $t_0\in I$.
+If $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma$ is differentiable curve in $G$ with $\gamma(t_0)=z_0$ for some $t_0\in I$.
 
 Then,
 
@@ -60,12 +60,14 @@ $$
 > 
 > $f$ is differentiable if and only if $f(z+h)=f(z)+f'(z)h+\frac{1}{2}h^2f''(z)+o(h^3)$ as $h\to 0$. (By Taylor expansion)
 
-Since $f$ is holomorphic at $\gamma(t_0)=\zeta_0$, we have
+Since $f$ is holomorphic at $\gamma(t_0)=z_0$, we have
 
 $$
-f(\zeta_0)=f(\zeta_0)+(\zeta-\zeta_0)f'(\zeta_0)+o(\zeta-\zeta_0)
+f(z_0)=f(z_0)+(z-z_0)f'(z_0)+o(z-z_0)
 $$
 
+> This result comes from Taylor Expansion of the derivative of the function around the point $z_0$
+
 and
 
 $$
@@ -85,7 +87,7 @@ $$
 \end{aligned}
 $$
 
-EOP
+QED
 
 #### Definition 2.12 (Conformal function)
 
@@ -93,7 +95,7 @@ A function $f:G\to \mathbb{C}$ is called conformal if it preserves the angle bet
 
 #### Theorem 2.13 (Conformal function)
 
-If $f:G\to \mathbb{C}$ is conformal at $\zeta_0\in G$, then $f$ is holomorphic at $\zeta_0$ and $f'(\zeta_0)\neq 0$.
+If $f:G\to \mathbb{C}$ is conformal at $z_0\in G$, then $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$.
 
 Example:
 
@@ -105,20 +107,20 @@ is not conformal at $z=0$ because $f'(0)=0$.
 
 #### Lemma of conformal function
 
-Suppose $f$ is real differentiable, let $a=\frac{\partial f}{\partial \zeta}(\zeta_0)$, $b=\frac{\partial f}{\partial \overline{\zeta}}(\zeta_0)$.
+Suppose $f$ is real differentiable, let $a=\frac{\partial f}{\partial z}(z_0)$, $b=\frac{\partial f}{\partial \overline{z}}(z_0)$.
 
-Let $\gamma(t_0)=\zeta_0$. Then $(f\circ \gamma)'(t_0)=a\gamma'(t_0)+b\overline{\gamma'(t_0)}$.
+Let $\gamma(t_0)=z_0$. Then $(f\circ \gamma)'(t_0)=a\gamma'(t_0)+b\overline{\gamma'(t_0)}$.
 
 Proof:
 
 $f=u+iv$, $u,v$ are real differentiable.
 
 $$
-a=\frac{\partial f}{\partial \zeta}=\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)
+a=\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)
 $$
 
 $$
-b=\frac{\partial f}{\partial \overline{\zeta}}=\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)
+b=\frac{\partial f}{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)
 $$
 
 $$
@@ -131,7 +133,7 @@ $$
 
 $$
 \begin{aligned}
-(f\circ \gamma)'(t_0)&=\frac{\partial f}{\partial \zeta}(\gamma(t_0))\gamma'(t_0)+\frac{\partial f}{\partial \overline{\zeta}}(\gamma(t_0))\overline{\gamma'(t_0)} \\
+(f\circ \gamma)'(t_0)&=\frac{\partial f}{\partial z}(\gamma(t_0))\gamma'(t_0)+\frac{\partial f}{\partial \overline{z}}(\gamma(t_0))\overline{\gamma'(t_0)} \\
 &=\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\right]\left(\frac{d\alpha}{dt}+i\frac{d\beta}{dt}\right)\\
 &+\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\right]\left(\frac{d\beta}{dt}-i\frac{d\alpha}{dt}\right) \\
 &=\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\frac{d\alpha}{dt}-\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\frac{d\beta}{dt}\right]\\
@@ -142,56 +144,56 @@ $$
 \end{aligned}
 $$
 
-EOP
+QED
 
 #### Theorem of differentiability
 
 Let $f:G\to \mathbb{C}$ be a function defined on an open set $G\subset \mathbb{C}$ that is both holomorphic and (real) differentiable, where $f=u+iv$ with $u,v$ real differentiable functions.
 
-Then, $f$ is conformal at every point $\zeta_0\in G$ if and only if $f$ is holomorphic at $\zeta_0$ and $f'(\zeta_0)\neq 0$.
+Then, $f$ is conformal at every point $z_0\in G$ if and only if $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$.
 
 Proof:
 
 We prove the equivalence in two parts.
 
-($\implies$) Suppose that $f$ is conformal at $\zeta_0$. By definition, conformality means that $f$ preserves angles (including their orientation) between any two intersecting curves through $\zeta_0$. In the language of real analysis, this requires that the (real) derivative (Jacobian) of $f$ at $\zeta_0$, $Df(\zeta_0)$, acts as a similarity transformation. Any similarity in $\mathbb{R}^2$ can be written as a rotation combined with a scaling; in particular, its matrix representation has the form
+($\implies$) Suppose that $f$ is conformal at $z_0$. By definition, conformality means that $f$ preserves angles (including their orientation) between any two intersecting curves through $z_0$. In the language of real analysis, this requires that the (real) derivative (Jacobian) of $f$ at $z_0$, $Df(z_0)$, acts as a similarity transformation. Any similarity in $\mathbb{R}^2$ can be written as a rotation combined with a scaling; in particular, its matrix representation has the form
 $$
 \begin{pmatrix}
 A & -B \\
 B & A
 \end{pmatrix},
 $$
-for some real numbers $A$ and $B$. This is exactly the matrix corresponding to multiplication by the complex number $a=A+iB$. Therefore, the Cauchy-Riemann equations must hold at $\zeta_0$, implying that $f$ is holomorphic at $\zeta_0$. Moreover, because the transformation is nondegenerate (preserving angles implies nonzero scaling), we must have $f'(\zeta_0)=a\neq 0$.
+for some real numbers $A$ and $B$. This is exactly the matrix corresponding to multiplication by the complex number $a=A+iB$. Therefore, the Cauchy-Riemann equations must hold at $z_0$, implying that $f$ is holomorphic at $z_0$. Moreover, because the transformation is nondegenerate (preserving angles implies nonzero scaling), we must have $f'(z_0)=a\neq 0$.
 
-($\impliedby$) Now suppose that $f$ is holomorphic at $\zeta_0$ and $f'(\zeta_0)\neq 0$. Then by the definition of the complex derivative, the first-order (linear) approximation of $f$ near $\zeta_0$ is
+($\impliedby$) Now suppose that $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$. Then by the definition of the complex derivative, the first-order (linear) approximation of $f$ near $z_0$ is
 $$
-f(\zeta_0+h)=f(\zeta_0)+f'(\zeta_0)h+o(|h|),
+f(z_0+h)=f(z_0)+f'(z_0)h+o(|h|),
 $$
-for small $h\in\mathbb{C}$. Multiplication by the nonzero complex number $f'(\zeta_0)$ is exactly a rotation and scaling (i.e., a similarity transformation). Therefore, for any smooth curve $\gamma(t)$ with $\gamma(t_0)=\zeta_0$, we have
+for small $h\in\mathbb{C}$. Multiplication by the nonzero complex number $f'(z_0)$ is exactly a rotation and scaling (i.e., a similarity transformation). Therefore, for any smooth curve $\gamma(t)$ with $\gamma(t_0)=z_0$, we have
 $$
-(f\circ\gamma)'(t_0)=f'(\zeta_0)\gamma'(t_0),
+(f\circ\gamma)'(t_0)=f'(z_0)\gamma'(t_0),
 $$
-and the angle between any two tangent vectors at $\zeta_0$ is preserved (up to the fixed rotation). Hence, $f$ is conformal at $\zeta_0$.
+and the angle between any two tangent vectors at $z_0$ is preserved (up to the fixed rotation). Hence, $f$ is conformal at $z_0$.
 
 For further illustration, consider the special case when $f$ is an affine map.
 
 Case 1: Suppose 
 $$
-f(\zeta)=a\zeta+b\overline{\zeta}.
+f(z)=az+b\overline{z}.
 $$
 The Wirtinger derivatives of $f$ are
 $$
-\frac{\partial f}{\partial \zeta}=a \quad \text{and} \quad \frac{\partial f}{\partial \overline{\zeta}}=b.
+\frac{\partial f}{\partial z}=a \quad \text{and} \quad \frac{\partial f}{\partial \overline{z}}=b.
 $$
-For $f$ to be holomorphic, we require $\frac{\partial f}{\partial \overline{\zeta}}=b=0$. Moreover, to have a nondegenerate (angle-preserving) map, we must have $a\neq 0$. If $b\neq 0$, then the map mixes $\zeta$ and $\overline{\zeta}$, and one can check that the linearization maps the real axis $\mathbb{R}$ into the set $\{(a+b)t\}$, which does not uniformly scale and rotate all directions. Thus, $f$ fails to be conformal when $b\neq 0$.
+For $f$ to be holomorphic, we require $\frac{\partial f}{\partial \overline{z}}=b=0$. Moreover, to have a nondegenerate (angle-preserving) map, we must have $a\neq 0$. If $b\neq 0$, then the map mixes $z$ and $\overline{z}$, and one can check that the linearization maps the real axis $\mathbb{R}$ into the set $\{(a+b)t\}$, which does not uniformly scale and rotate all directions. Thus, $f$ fails to be conformal when $b\neq 0$.
 
 Case 2: For a general holomorphic function, the lemma of conformal functions shows that if 
 $$
-(f\circ \gamma)'(t_0)=f'(\zeta_0)\gamma'(t_0)
+(f\circ \gamma)'(t_0)=f'(z_0)\gamma'(t_0)
 $$
-for any differentiable curve $\gamma$ through $\zeta_0$, then the effect of $f$ near $\zeta_0$ is exactly given by multiplication by $f'(\zeta_0)$. Since multiplication by a nonzero complex number is a similarity transformation, $f$ is conformal at $\zeta_0$.
+for any differentiable curve $\gamma$ through $z_0$, then the effect of $f$ near $z_0$ is exactly given by multiplication by $f'(z_0)$. Since multiplication by a nonzero complex number is a similarity transformation, $f$ is conformal at $z_0$.
 
-EOP
+QED
 
 ### Harmonic function
 
@@ -227,7 +229,7 @@ $$
 \Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 v}{\partial x\partial y}-\frac{\partial^2 v}{\partial y\partial x}=0.
 $$
 
-EOP
+QED
 
 If $v$ is such that $f=u+iv$ is holomorphic on $\Omega$, then $v$ is called harmonic conjugate of $u$ on $\Omega$.
 

pages/Math416/Math416_L5.md

@@ -18,14 +18,14 @@ So
 
 $$
 \begin{aligned}
-\frac{\partial f}{\partial \zeta}&=\frac{1}{2}\left(u_x+v_y\right)-i\frac{1}{2}\left(v_x+u_y\right)\\
+\frac{\partial f}{\partial z}&=\frac{1}{2}\left(u_x+v_y\right)-i\frac{1}{2}\left(v_x+u_y\right)\\
 &=\frac{1}{2}\left(\alpha+\delta\right)-i\frac{1}{2}\left(\beta-\sigma\right)\\
 \end{aligned}
 $$
 
 $$
 \begin{aligned}
-\frac{\partial f}{\partial \overline{\zeta}}&=\frac{1}{2}\left(u_x+v_y\right)+i\frac{1}{2}\left(v_x+u_y\right)\\
+\frac{\partial f}{\partial \overline{z}}&=\frac{1}{2}\left(u_x+v_y\right)+i\frac{1}{2}\left(v_x+u_y\right)\\
 &=\frac{1}{2}\left(\alpha-\delta\right)+i\frac{1}{2}\left(\beta+\sigma\right)\\
 \end{aligned}
 $$
@@ -42,11 +42,11 @@ $$
 So,
 
 $$
-\frac{\partial f}{\partial \zeta}=\frac{1}{2}(\alpha+\alpha)+i\frac{1}{2}(\beta+\beta)=a
+\frac{\partial f}{\partial z}=\frac{1}{2}(\alpha+\alpha)+i\frac{1}{2}(\beta+\beta)=a
 $$
 
 $$
-\frac{\partial f}{\partial \overline{\zeta}}=\frac{1}{2}(\alpha-\alpha)+i\frac{1}{2}(\beta-\beta)=0
+\frac{\partial f}{\partial \overline{z}}=\frac{1}{2}(\alpha-\alpha)+i\frac{1}{2}(\beta-\beta)=0
 $$
 
 > Less pain to represent a complex function using four real numbers.
@@ -59,10 +59,10 @@ Let $a,b,c,d$ be complex numbers. such that $ad-bc\neq 0$.
 The linear fractional transformation is defined as
 
 $$
-\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}
+\phi(z)=\frac{az+b}{cz+d}
 $$
 
-If we let $\psi(\zeta)=\frac{e\zeta-f}{-g\zeta+h}$ also be a linear fractional transformation, then $\phi\circ\psi$ is also a linear fractional transformation.
+If we let $\psi(z)=\frac{ez-f}{-gz+h}$ also be a linear fractional transformation, then $\phi\circ\psi$ is also a linear fractional transformation.
 
 New coefficients can be solved by
 
@@ -82,7 +82,7 @@ m&n
 \end{pmatrix}
 $$
 
-So $\phi\circ\psi(\zeta)=\frac{k\zeta+l}{m\zeta+n}$
+So $\phi\circ\psi(z)=\frac{kz+l}{mz+n}$
 
 ### Complex projective space
 
@@ -98,7 +98,7 @@ $\mathbb{C}P^1$ is the set of lines through the origin in $\mathbb{C}$.
 
 We defined $(a,b)\sim(c,d),(a,b),(c,d)\in\mathbb{C}\setminus\{(0,0)\}$ if $\exists t\neq 0,t\in\mathbb{C}\setminus\{0\}$ such that $(a,b)=(tc,td)$.
 
-So, $\forall \zeta\in\mathbb{C}\setminus\{0\}$:
+So, $\forall z\in\mathbb{C}\setminus\{0\}$:
 
 If $a\neq 0$, then $(a,b)\sim(1,\frac{b}{a})$.
 
@@ -116,33 +116,33 @@ c & d
 Suppose $M$ is non-singular. Then $ad-bc\neq 0$.
 
 If $M\begin{pmatrix}
-\zeta_1\\
-\zeta_2
+z_1\\
+z_2
 \end{pmatrix}=\begin{pmatrix}
 \omega_1\\
 \omega_2
 \end{pmatrix}$, then $M\begin{pmatrix}
-t\zeta_1\\
-t\zeta_2
+tz_1\\
+tz_2
 \end{pmatrix}=\begin{pmatrix}
 t\omega_1\\
 t\omega_2
 \end{pmatrix}$.
 
 So, $M$ induces a map $\phi_M:\mathbb{C}P^1\to\mathbb{C}P^1$ defined by $M\begin{pmatrix}
-\zeta\\
+z\\
 1
 \end{pmatrix}=\begin{pmatrix}
-\frac{a\zeta+b}{c\zeta+d}\\
+\frac{az+b}{cz+d}\\
 1
 \end{pmatrix}$.
 
-$\phi_M(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
+$\phi_M(z)=\frac{az+b}{cz+d}$.
 
 If we let $M_2=\begin{pmatrix}
 e &f\\
 g &h
-\end{pmatrix}$, where $ad-bc\neq 0$ and $eh-fg\neq 0$, then $\phi_{M_2}(\zeta)=\frac{e\zeta+f}{g\zeta+h}$.
+\end{pmatrix}$, where $ad-bc\neq 0$ and $eh-fg\neq 0$, then $\phi_{M_2}(z)=\frac{ez+f}{gz+h}$.
 
 So, $M_2M_1=\begin{pmatrix}
 a&b\\
@@ -151,15 +151,15 @@ c&d
 e&f\\
 g&h
 \end{pmatrix}=\begin{pmatrix}
-\zeta\\
+z\\
 1
 \end{pmatrix}$.
 
 This also gives $\begin{pmatrix}
-k\zeta+l\\
-m\zeta+n
+kz+l\\
+mz+n
 \end{pmatrix}\sim\begin{pmatrix}
-\frac{k\zeta+l}{m\zeta+n}\\
+\frac{kz+l}{mz+n}\\
 1
 \end{pmatrix}$.
 
@@ -187,17 +187,17 @@ If $\phi$ is a non-constant linear fractional transformation, then $\phi$ is con
 
 Proof:
 
-Know that $\phi_0\circ\phi(\zeta)=\zeta$,
+Know that $\phi_0\circ\phi(z)=z$,
 
-Then $\phi(\zeta)=\phi_0^{-1}\circ\phi\circ\phi_0(\zeta)$.
+Then $\phi(z)=\phi_0^{-1}\circ\phi\circ\phi_0(z)$.
 
-So $\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
+So $\phi(z)=\frac{az+b}{cz+d}$.
 
 $\phi:\mathbb{C}\cup\{\infty\}\to\mathbb{C}\cup\{\infty\}$ which gives $\phi(\infty)=\frac{a}{c}$ and $\phi(-\frac{d}{c})=\infty$.
 
 So, $\phi$ is conformal.
 
-EOP
+QED
 
 #### Proposition 3.4 of Fixed points
 
@@ -205,7 +205,7 @@ Any non-constant linear fractional transformation except the identity transforma
 
 Proof:
 
-Let $\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
+Let $\phi(z)=\frac{az+b}{cz+d}$.
 
 Case 1: $c=0$
 
@@ -213,19 +213,19 @@ Then $\infty$ is a fixed point.
 
 Case 2: $c\neq 0$
 
-Then $\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
+Then $\phi(z)=\frac{az+b}{cz+d}$.
 
-The solution of $\phi(\zeta)=\zeta$ is $c\zeta^2+(d-a)\zeta-b=0$.
+The solution of $\phi(z)=z$ is $cz^2+(d-a)z-b=0$.
 
-Such solutions are $\zeta=\frac{-(d-a)\pm\sqrt{(d-a)^2+4bc}}{2c}$.
+Such solutions are $z=\frac{-(d-a)\pm\sqrt{(d-a)^2+4bc}}{2c}$.
 
 So, $\phi$ has 1 or 2 fixed points.
 
-EOP
+QED
 
 #### Proposition 3.5 of triple transitivity
 
-If $\zeta_1,\zeta_2,\zeta_3\in\mathbb{C}P^1$ are distinct, then there exists a non-constant linear fractional transformation $\phi$ such that $\phi(\zeta_1)=\zeta_2$ and $\phi(\zeta_3)=\infty$.
+If $z_1,z_2,z_3\in\mathbb{C}P^1$ are distinct, then there exists a non-constant linear fractional transformation $\phi$ such that $\phi(z_1)=z_2$ and $\phi(z_3)=\infty$.
 
 Proof as homework.
 
@@ -235,6 +235,4 @@ We defined clircle to be a circle or a line.
 
 If $\phi$ is a non-constant linear fractional transformation, then $\phi$ maps clircles to clircles.
 
-Proof:
-
-Continue on next lecture.
+Proof continue on next lecture.

pages/Math416/Math416_L6.md

@@ -12,16 +12,16 @@ We defined clircle to be a circle or a line.
 
 The circle equation is:
 
-Let $\zeta=u+iv$ be the center of the circle, $r$ be the radius of the circle.
+Let $z=u+iv$ be the center of the circle, $r$ be the radius of the circle.
 
 $$
-circle=\{z\in\mathbb{C}:|\zeta-c|=r\}
+circle=\{z\in\mathbb{C}:|z-c|=r\}
 $$
 
 This is:
 
 $$
-|\zeta|^2-c\overline{\zeta}-\overline{c}\zeta+|c|^2-r^2=0
+|z|^2-c\overline{z}-\overline{c}z+|c|^2-r^2=0
 $$
 
 If $\phi$ is a non-constant linear fractional transformation, then $\phi$ maps clircles to clircles.
@@ -29,7 +29,7 @@ If $\phi$ is a non-constant linear fractional transformation, then $\phi$ maps c
 We claim that a map is circle preserving if and only if for some $\alpha,\beta,\gamma,\delta\in\mathbb{R}$.
 
 $$
-\alpha|\zeta|^2+\beta Re(\zeta)+\gamma Im(\zeta)+\delta=0
+\alpha|z|^2+\beta Re(z)+\gamma Im(z)+\delta=0
 $$
 
 when $\alpha=0$, it is a line.
@@ -38,7 +38,7 @@ when $\alpha\neq 0$, it is a circle.
 
 Proof:
 
-Let $w=u+iv=\frac{1}{\zeta}$, so $\frac{1}{w}=\frac{u}{u^2+v^2}-i\frac{v}{u^2+v^2}$.
+Let $w=u+iv=\frac{1}{z}$, so $\frac{1}{w}=\frac{u}{u^2+v^2}-i\frac{v}{u^2+v^2}$.
 
 Then the original equation becomes:
 
@@ -48,13 +48,13 @@ $$
 
 Which is in the form of circle equation.
 
-EOP
+QED
 
 ## Chapter 4 Elementary functions
 
 > $e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}$
 
-So, following the definition of $e^\zeta$, we have:
+So, following the definition of $e^z$, we have:
 
 $$
 \begin{aligned}
@@ -65,51 +65,51 @@ e^{x+iy}&=e^xe^{iy} \\
 \end{aligned}
 $$
 
-### $e^\zeta$
+### $e^z$
 
-The exponential of $e^\zeta=x+iy$ is defined as:
+The exponential of $e^z=x+iy$ is defined as:
 
 $$
-e^\zeta=exp(\zeta)=e^x(\cos y+i\sin y)
+e^z=exp(z)=e^x(\cos y+i\sin y)
 $$
 
 So,
 
 $$
-|e^\zeta|=|e^x||\cos y+i\sin y|=e^x
+|e^z|=|e^x||\cos y+i\sin y|=e^x
 $$
 
-#### Theorem 4.3 $e^\zeta$ is holomorphic
+#### Theorem 4.3 $e^z$ is holomorphic
 
-$e^\zeta$ is holomorphic on $\mathbb{C}$.
+$e^z$ is holomorphic on $\mathbb{C}$.
 
 Proof:
 
 $$
 \begin{aligned}
-\frac{\partial}{\partial\zeta}e^\zeta&=\frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{i}{\partial y}\right)e^x(\cos y+i\sin y) \\
+\frac{\partial}{\partial z}e^z&=\frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{i}{\partial y}\right)e^x(\cos y+i\sin y) \\
 &=\frac{1}{2}e^x(\cos y+i\sin y)+ie^x(-\sin y+i\cos y) \\
 &=0
 \end{aligned}
 $$
 
-EOP
+QED
 
-#### Theorem 4.4 $e^\zeta$ is periodic
+#### Theorem 4.4 $e^z$ is periodic
 
-$e^\zeta$ is periodic with period $2\pi i$.
+$e^z$ is periodic with period $2\pi i$.
 
 Proof:
 
 $$
-e^{\zeta+2\pi i}=e^\zeta e^{2\pi i}=e^\zeta\cdot 1=e^\zeta
+e^{z+2\pi i}=e^z e^{2\pi i}=e^z\cdot 1=e^z
 $$
 
-EOP
+QED
 
-#### Theorem 4.5 $e^\zeta$ as a map
+#### Theorem 4.5 $e^z$ as a map
 
-$e^\zeta$ is a map from $\mathbb{C}$ to $\mathbb{C}$ with period $2\pi i$.
+$e^z$ is a map from $\mathbb{C}$ to $\mathbb{C}$ with period $2\pi i$.
 
 $$
 e^{\pi i}+1=0
@@ -119,36 +119,36 @@ This is a map from cartesian coordinates to polar coordinates, where $e^x$ is th
 
 This map attains every value in $\mathbb{C}\setminus\{0\}$.
 
-#### Definition 4.6-8 $\cos\zeta$ and $\sin\zeta$
+#### Definition 4.6-8 $\cos z$ and $\sin z$
 
 $$
-\cos\zeta=\frac{1}{2}(e^{i\zeta}+e^{-i\zeta})
+\cos z=\frac{1}{2}(e^{iz}+e^{-iz})
 $$
 
 $$
-\sin\zeta=\frac{1}{2i}(e^{i\zeta}-e^{-i\zeta})
+\sin z=\frac{1}{2i}(e^{iz}-e^{-iz})
 $$
 
 $$
-\cosh\zeta=\frac{1}{2}(e^\zeta+e^{-\zeta})
+\cosh z=\frac{1}{2}(e^z+e^{-z})
 $$
 
 $$
-\sinh\zeta=\frac{1}{2}(e^\zeta-e^{-\zeta})
+\sinh z=\frac{1}{2}(e^z-e^{-z})
 $$
 
-From this definition, we can see that $\cos\zeta$ and $\sin\zeta$ are no longer bounded.
+From this definition, we can see that $\cos z$ and $\sin z$ are no longer bounded in the complex plane.
 
-And this definition is still compatible with the previous definition of $\cos$ and $\sin$ when $\zeta$ is real.
+And this definition is still compatible with the previous definition of $\cos$ and $\sin$ when $z$ is real.
 
 Moreover,
 
 $$
-\cosh(i\zeta)=\cos\zeta
+\cosh(iz)=\cos z
 $$
 
 $$
-\sinh(i\zeta)=i\sin\zeta
+\sinh(iz)=i\sin z
 $$
 
 ### Logarithm
@@ -175,37 +175,58 @@ If $y\in(-\pi,\pi]$, then $\log a=b$ means $e^b=a$ and $Im(b)\in(-\pi,\pi]$.
 
 If $a=re^{i\theta}$, then $\log a=\log r+i(\theta_0+2k\pi)$.
 
-#### Definition 4.10
+#### Definition 4.10 of Branch of $\arg z$ and $\log z$
 
 Let $G$ be an open connected subset of $\mathbb{C}\setminus\{0\}$.
 
-A branch of $\arg(\zeta)$ in $G$ is a continuous function $\alpha$, such that $\alpha(\zeta)$ is a value of $\arg(\zeta)$.
+A branch of $\arg(z)$ in $G$ is a continuous function $\alpha:G\to G$, such that $\alpha(z)$ is a value of $\arg(z)$.
 
-A branch of $\log(\zeta)$ in $G$ is a continuous function $\beta$, such that $e^{\beta(\zeta)}=\zeta$.
+A branch of $\log(z)$ in $G$ is a continuous function $\beta$, such that $e^{\beta(z)}=z$.
 
-Note: $G$ has a branch of $\arg(\zeta)$ if and only if it has a branch of $\log(\zeta)$.
+Note: $G$ has a branch of $\arg(z)$ if and only if it has a branch of $\log(z)$.
 
-If $G=\mathbb{C}\setminus\{0\}$, then  not branch of $\arg(\zeta)$ exists.
+Proof:
 
-Suppose $\alpha_1$ and $\alpha_2$ are two branches of $\arg(\zeta)$ in $G$.
+Suppose there exists $\alpha(z)$ such that $\forall z\in G$, $\alpha(z)\in G$, then $l(z)=\ln|z|+i\alpha(z)$ is a branch of $\log(z)$.
+
+Suppose there exists $l(z)$ such that $\forall z\in G$, $l(z)\in G$, then $\alpha(z)=Im(z)$ is a branch of $\arg(z)$.
+
+QED
+
+If $G=\mathbb{C}\setminus\{0\}$, then  not branch of $\arg(z)$ exists.
+
+#### Corollary of 4.10
+
+Suppose $\alpha_1$ and $\alpha_2$ are two branches of $\arg(z)$ in $G$.
 
 Then,
 
 $$
-\alpha_1(\zeta)-\alpha_2(\zeta)=2k\pi i
+\alpha_1(z)-\alpha_2(z)=2k\pi
+$$
+
+for some $k\in\mathbb{Z}$.
+
+
+Suppose $l_1$ and $l_2$ are two branches of $\log(z)$ in $G$.
+
+Then,
+
+$$
+l_1(z)-l_2(z)=2k\pi i
 $$
 
 for some $k\in\mathbb{Z}$.
 
 #### Theorem 4.11
 
-$\log(\zeta)$ is holomorphic on $\mathbb{C}\setminus\{0\}$.
+$\log(z)$ is holomorphic on $\mathbb{C}\setminus\{0\}$.
 
 Proof:
 
 Method 1: Use polar coordinates. (See in homework)
 
-Method 2: Use the fact that $\log(\zeta)$ is the inverse of $e^\zeta$.
+Method 2: Use the fact that $\log(z)$ is the inverse of $e^z$.
 
 Suppose $h=s+it$, $e^h=e^s(\cos t+i\sin t)$, $e^h-1=e^s(\cos t-1)+i\sin t$. So
 

pages/Math416/Math416_L7.md

@@ -8,7 +8,7 @@ $$
 e^z=e^{x+iy}=e^x(\cos y+i\sin y)
 $$
 
-### Logarithm
+### Logarithm Reviews
 
 #### Definition 4.9 Logarithm
 
@@ -24,28 +24,52 @@ A branch of logarithm is a continuous function $f$ on a domain $D$ such that $e^
 
 #### Theorem 4.11
 
-$\log(\zeta)$ is holomorphic on $\mathbb{C}\setminus\{0\}$.
+$\log(z)$ is holomorphic on $\mathbb{C}\setminus\{0\}$.
 
 Proof:
 
-We proved that $\frac{\partial}{\partial\overline{z}}e^{\zeta}=0$ on $\mathbb{C}\setminus\{0\}$.
+We proved that $\frac{\partial}{\partial\overline{z}}e^{z}=0$ 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.
+Then $\frac{d}{dz}e^{z}=\frac{\partial}{\partial x}e^{z}=0$ if we know that $e^{z}$ 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 $\frac{d}{dz}e^{z}=e^{z}$, we know that $e^{z}$ 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 $\frac{d}{dz}\log(\zeta)=\frac{1}{e^{\log(\zeta)}}=\frac{1}{\zeta}$.
+Since $\exp(\log(z))=z$, we know that $\log(z)$ is the inverse of $\exp(z)$, so $\frac{d}{dz}\log(z)=\frac{1}{e^{\log(z)}}=\frac{1}{z}$.
 
-EOP
+QED
 
 We call $\frac{f'}{f}$ the logarithmic derivative of $f$.
 
+#### Definition 4.16
+
+_I don't know if this material is covered or not, so I will add it here to prevent confusion for future readers_
+
+If $a$ and $c$ are complex numbers, with $a\neq 0$, then by the values of $a^c$ one means the value of $e^{c\log a}$.
+
+For example, $1^i=e^{i (2\pi n i)}$
+
+If you accidentally continue on this section and find it interesting, you will find Riemann zeta function
+
+$$
+z(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}
+$$
+
+And analytic continuation for such function for number less than or equal to $1$.
+
+And perhaps find trivial zeros for negative integers on real line. It is important to note that the Riemann zeta function has non-trivial zeros, which are located in the critical strip where the real part of $s$ is between 0 and 1. The famous Riemann Hypothesis conjectures that all non-trivial zeros lie on the critical line where the real part of $s$ is $\frac{1}{2}$.
+
 ## Chapter 5. Power series
 
+### Convergence
+
+#### Necessary Condition for Convergence
+
 If $\sum_{n=0}^{\infty}c_n$ converges, then $\lim_{n\to\infty}c_n=0$ exists.
 
 ### Geometric series
 
+Let $c$ be a complex number
+
 $$
 \sum_{n=0}^{N}c^n=\frac{1-c^{N+1}}{1-c}
 $$
@@ -66,11 +90,13 @@ If $|c|<1$, then $\lim_{N\to\infty}c^{N+1}=0$, so $\lim_{N\to\infty}(1-c)(1+c+c^
 
 If $|c|\geq 1$, then $c^{N+1}$ does not converge to 0, so the series diverges.
 
-EOP
+QED
 
-### Convergence
+#### Theorem 5.4 (Triangle Inequality for Series)
 
-#### Definition 5.4
+If the series $\sum_{n=0}^{\infty}c_n$ converges, then $\left|\sum_{n=0}^\infty c_n\right|\leq \sum_{n=0}^{\infty}|c_n|$.
+
+#### Definition 5.5
 
 $$
 \sum_{n=0}^{\infty}c_n
@@ -96,57 +122,61 @@ A sequence of functions $f_n$ **converges locally uniformly** to $f$ on a set $G
 
 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.?
+#### Theorem 5.7
 
-If the subsequence of a converging sequence of functions converges (a), then the original sequence converges (a).
+If the subsequence (or partial sum) of a converging sequence of functions converges (a), then the original sequence converges (a).
+
+The N-th partial sum of the series $\sum_{n=0}^\infty f_n$ is $\sum_{n=0}^{N}f_n$
 
 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 $a^b=e^{b\log a}\{e^{2k\pi ik b}\},k\in\mathbb{Z}$.
+> Corollary from definition of $a^b$ in complex plane
+> 
+> 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 $\sum_{n=0}^{\infty}c_n(\zeta-\zeta_0)^n$.
+A power series is a series of the form $\sum_{n=0}^{\infty}c_n(z-z_0)^n$.
 
-#### Theorem 5.10
+#### Definition 5.9 Region of Convergence
 
-For every power series, there exists a radius of convergence $R$ such that the series converges absolutely and locally uniformly on $B(\zeta_0,R)$.
+For every power series, there exists a radius of convergence $r$ such that the series converges absolutely and locally uniformly on $B_r(z_0)$.
 
-And it diverges pointwise outside $B(\zeta_0,R)$.
+And it diverges pointwise outside $B_r(z_0)$.
 
 Proof:
 
-Without loss of generality, we can assume that $\zeta_0=0$.
+Without loss of generality, we can assume that $z_0=0$.
 
-Suppose that the power series is $\sum_{n=0}^{\infty}c_n (\zeta)^n$ converges at $\zeta=re^{i\theta}$.
+Suppose that the power series is $\sum_{n=0}^{\infty}c_n (z)^n$ converges at $z=re^{i\theta}$.
 
-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 want to show that the series converges absolutely and uniformly on $\overline{B_r(0)}$ (_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 $\forall \zeta\in\overline{B(0,r)}$, $|c_n\zeta^n|\leq |c_n| |\zeta|^n \leq M \left(\frac{|\zeta|}{r}\right)^n$.
+So $\forall z\in\overline{B_r(0)}$, $|c_nz^n|\leq |c_n| |z|^n \leq M \left(\frac{|z|}{r}\right)^n$.
 
-So $\sum_{n=0}^{\infty}|c_n\zeta^n|$ converges absolutely.
+So $\sum_{n=0}^{\infty}|c_nz^n|$ converges absolutely.
 
-So the series converges absolutely and uniformly on $\overline{B(0,r)}$.
+So the series converges absolutely and uniformly on $\overline{B_r(0)}$.
 
-If $|\zeta| > r$, then $|c_n \zeta^n|$ does not tend to zero, and the series diverges.
+If $|z| > r$, then $|c_n z^n|$ does not tend to zero, and the series diverges.
 
-EOP
+QED
+
+We denote this $r$ captialized by te radius of convergence
 
 #### Possible Cases for the Convergence of Power Series
 
-1. **Convergence Only at $\zeta = 0$**:
-   - **Proof**: If the power series $\sum_{n=0}^{\infty} c_n (\zeta - \zeta_0)^n$ converges only at $\zeta = 0$, it means that the radius of convergence $R = 0$. This occurs when the terms $c_n (\zeta - \zeta_0)^n$ do not tend to zero for any $\zeta \neq 0$. The series diverges for all $\zeta \neq 0$ because the terms grow without bound.
+1. **Convergence Only at $z = 0$**:
+   - **Proof**: If the power series $\sum_{n=0}^{\infty} c_n (z - z_0)^n$ converges only at $z = 0$, it means that the radius of convergence $R = 0$. This occurs when the terms $c_n (z - z_0)^n$ do not tend to zero for any $z \neq 0$. The series diverges for all $z \neq 0$ because the terms grow without bound.
 
 2. **Convergence Everywhere**:
-   - **Proof**: If the power series converges for all $\zeta \in \mathbb{C}$, the radius of convergence $R = \infty$. This implies that the terms $c_n (\zeta - \zeta_0)^n$ tend to zero for all $\zeta$. This can happen if the coefficients $c_n$ decrease rapidly enough, such as in the exponential series.
+   - **Proof**: If the power series converges for all $z \in \mathbb{C}$, the radius of convergence $R = \infty$. This implies that the terms $c_n (z - z_0)^n$ tend to zero for all $z$. This can happen if the coefficients $c_n$ decrease rapidly enough, such as in the exponential series.
 
 3. **Convergence Within a Finite Radius**:
-   - **Proof**: For a power series with a finite radius of convergence $R$, the series converges absolutely and uniformly for $|\zeta - \zeta_0| < R$ and diverges for $|\zeta - \zeta_0| > R$. On the boundary $|\zeta - \zeta_0| = R$, the series may converge or diverge depending on the specific series. This is determined by the behavior of the terms on the boundary.
+   - **Proof**: For a power series with a finite radius of convergence $R$, the series converges absolutely and uniformly for $|z - z_0| < R$ and diverges for $|z - z_0| > R$. On the boundary $|z - z_0| = R$, the series may converge or diverge depending on the specific series. This is determined by the behavior of the terms on the boundary.

pages/Math416/Math416_L8.md

@@ -10,47 +10,58 @@ Let $f_n: G \to \mathbb{C}$ be a sequence of functions.
 
 Definition:
 
-Let $\zeta\in G$, $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $|f_n(\zeta) - f(\zeta)| < \epsilon$.
+Let $z\in G$, $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $|f_n(z) - f(z)| < \epsilon$.
 
 #### Convergence Uniformly
 
 Definition:
 
-$\forall \epsilon > 0$, $\forall \zeta\in G$, $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $|f_n(\zeta) - f(\zeta)| < \epsilon$.
+$\forall \epsilon > 0$, $\forall z\in G$, $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $|f_n(z) - f(z)| < \epsilon$.
 
 #### Convergence Locally Uniformly
 
 Definition:
 
-$\forall \epsilon > 0$, $\forall \zeta\in G$, $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $|f_n(\zeta) - f(\zeta)| < \epsilon$.
+$\forall \epsilon > 0$, $\forall z\in G$, $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $|f_n(z) - f(z)| < \epsilon$.
 
 #### Convergence Uniformly on Compact Sets
 
-Definition: $\forall C\subset G$ that is compact, $\forall \epsilon > 0, \exists N \in \mathbb{N} \text{ s.t. } \forall n \geq N, \forall \zeta\in C, |f_n(\zeta) - f(\zeta)| < \epsilon$
+Definition: $\forall C\subset G$ that is compact, $\forall \epsilon > 0, \exists N \in \mathbb{N} \text{ s.t. } \forall n \geq N, \forall z\in C, |f_n(z) - f(z)| < \epsilon$
 
 #### Power Series
 
 Definition:
 
 $$
-\sum_{n=0}^{\infty} c_n (\zeta - \zeta_0)^n
+\sum_{n=0}^{\infty} c_n (z - z_0)^n
 $$
 
-$\zeta_0$ is the center of the power series.
+$z_0$ is the center of the power series.
 
-#### Theorem of Power Seriess
+#### Theorem of Power Series
 
-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.
+If a power series converges at $z_0$, then it converges absolutely at every point of $\overline{B_r(z_0)}$ that is strictly inside the disk of convergence.
 
 ## Continue on Power Series
 
+### Review on $\limsup$
+
+The $\limsup(a_n)$ $a_n\in\mathbb{R}$ is defined as the sup of subsequence of $(a_n)$ as $n$ approaches infinity.
+
+It has the following properties that is useful for proving the remaining parts for this course.
+
+Suppose $(a_n)_1^\infty$ is a sequence of real numbers
+
+1. If $\rho\in \mathbb{R}$ satisfies that $\rho<\limsup_{n\to\infty}a_n$, then $\{a_n : a_n > \rho\}$ is infinite.
+2. If $\rho\in \mathbb{R}$ satisfies that $\rho>\limsup_{n\to\infty}a_n$, then $\{a_n : a_n > \rho\}$ is finite.
+
 ### 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
+The radius of convergence of the power series is given by $\sum_{n=0}^{\infty} a_n (z - z_0)^n$ is given by
 
 $$
 \frac{1}{R} = \limsup_{n\to\infty} |a_n|^{1/n}
@@ -74,52 +85,98 @@ 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 $\frac{1}{\limsup_{n\to\infty} |a_n|^{1/n}}$.
+We want to show that the radius of convergence of $\sum_{n=0}^{\infty} a_n (z - z_0)^n$ 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 $\zeta_0$.
+Since $\sum_{n=0}^{\infty} 1z^n=\frac{1}{1-z}$ for $|z|<1$. Assume $\limsup_{n\to\infty} |a_n|^{1/n}$ is finite, then $\sum_{n=0}^{\infty} a_n (z - z_0)^n$ converges absolutely at $z_0$.
 
-Let $\rho>\limsup_{n\to\infty} |a_n|^{1/n}$, then $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $|a_n|^{1/n}\leq \rho$.
+Let $\rho>\limsup_{n\to\infty} |a_n|^{1/n}$, then $\exists N \in \mathbb{N}$ such that $\forall n \geq N$, $|a_n|^{1/n}\leq \rho$. (By property of $\limsup$)
 
-So $\frac{1}{R}=\limsup_{n\to\infty} |a_n|^{1/n}<\rho$
+So $\frac{1}{R}=\limsup_{n\to\infty} |a_n|^{1/n}\leq\rho$
 
-So $R>\frac{1}{\rho}$
-
-/*TRACK LOST*/
+So $R\geq\frac{1}{\rho}$
 
 Backward direction:
 
-Suppose $|\zeta|>R$, then $\exists$ number $|\zeta|$ such that $|\zeta|>\frac{1}{\rho}>R$.
+Suppose $|z|>R$, then $\exists$ number $|z|$ such that $|z|>\frac{1}{\rho}\geq R$.
 
 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 $|a_{n_j}\zeta^{n_j}|>\rho^{n_j}|\zeta|^{n_j}$
+So $|a_{n_j}z^{n_j}|>\rho^{n_j}|z|^{n_j}$
 
-Series $\sum_{n=1}^{\infty} a_n\zeta^n$ diverges, each individual term is not going to $0$.
+Series $\sum_{n=1}^{\infty} a_nz^n$ diverges, each individual term is not going to $0$.
 
-So $\sum_{n=0}^{\infty} a_n (\zeta - \zeta_0)^n$ does not converge at $\zeta$
+So $\sum_{n=0}^{\infty} a_n (z - z_0)^n$ does not converge at $z$ if $|z|> \frac{1}{\rho}\geq R$
 
-EOP
+So $R=\frac{1}{\rho}$.
 
-_What if $|\zeta-\zeta_0|=R$?_
+QED
 
-For $\sum_{n=0}^{\infty} \zeta^n$, the radius of convergence is $1$.
+_What if $|z-z_0|=R$?_
+
+For $\sum_{n=0}^{\infty} z^n$, 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 $1$.
+For $\sum_{n=0}^{\infty} \frac{1}{(n+1)^2}z^n$, 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 $1$.
+For $\sum_{n=0}^{\infty} \frac{1}{n+1}z^n$, the radius of convergence is $1$.
 
-This diverges at $\zeta=1$ (harmonic series) and converges at $\zeta=-1$ (alternating harmonic series).
+This diverges at $z=1$ (harmonic series) and converges at $z=-1$ (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 $f'(\zeta)=\sum_{n=1}^{\infty} n a_n (\zeta - \zeta_0)^{n-1}=\sum_{k=0}^{\infty} (k+1)a_{k+1} (\zeta - \zeta_0)^k$.
+Differentiation of power series
+
+Suppose $\sum_{n=0}^{\infty} a_n (z - z_0)^n$ has a positive radius of convergence $R$. Define $f(z)=\sum_{n=0}^{\infty} a_n (z - z_0)^n$, then $f$ is holomorphic on $B_R(0)$ and $f'(z)=\sum_{n=1}^{\infty} n a_n (z - z_0)^{n-1}=\sum_{k=0}^{\infty} (k+1)a_{k+1} (z - z_0)^k$.
+
+> Here below is the proof on book, which will be covered in next lecture.
 
 Proof:
 
-/*TRACK LOST*/
+Without loss of generality, assume $z_0=0$. Let $R$ be the radius of convergence for the two power series: $\sum_{n=0}^{\infty} a_n z^n$ and $\sum_{n=1}^{\infty} n a_n z ^{n-1}$. The two power series have the same radius of convergence $|R|$.
+
+> For $z,w\in \mathbb{C}, n\in \N$, $$z^n-w^n=(z-w)\sum_{k=0}^{n-1} z^k w^{n-k-1}$$
+
+Let $z_1\in B_R(0)$, $|z_1|<\rho<R$ for some $\rho\in\mathbb{R}$.
+
+$$
+\begin{aligned}
+\frac{f(z)-f(z_1)}{z-z_1}-g(z_1)&=\frac{1}{z-z_1}\left[\sum_{n=0}^\infty a_n z^n -\sum_{n=0}^\infty a_n z_1^n\right]-\sum_{n=1}^{\infty} n a_n z_1 ^{n-1}\\
+&=\sum_{n=1}^{\infty} a_n \left[\frac{z^n-z_1^n}{z-z_1}-nz_1^{n-1}\right]\\
+&=\sum_{n=1}^{\infty} a_n \left[\left(\sum_{k=0}^{n-1}z^kz_1^{n-k-1}\right)-nz_1^{n-1}\right]\\
+&=\sum_{n=2}^{\infty} a_n \left[\sum_{k=1}^{n-1}z_1^{n-k-1}(z^k-z^k_1)\right]
+\end{aligned}
+$$
+
+Using the lemma again we get
+
+$$
+\begin{aligned}
+|z^k-z_1^k|&=|z-z_1|\left|\sum_{j=0}^{k-1}z_jz_1^{k-j-1}\right|\\
+&\leq |z-z_1| \sum_{j=0}^{k-1}|z_j||z_1^{k-j-1}|\\
+&\leq k\rho^{k-1}|z-z_1|
+\end{aligned}
+$$
+
+Then,
+
+$$
+\begin{aligned}
+\left|\frac{f(z)-f(z_1)}{z-z_1}-g(z_1)\right|&=\left|\sum_{n=2}^{\infty} a_n \left[\sum_{k=1}^{n-1}z_1^{n-k-1}(z^k-z^k_1)\right]\right|\\
+&\leq \sum_{n=2}^{\infty} |a_n| \left[\sum_{k=1}^{n-1}|z_1|^{n-k-1}|z^k-z_1^k|\right]\\
+&\leq \sum_{n=2}^{\infty} |a_n|  \left[ \sum_{k=1}^{n-1} \rho^{n-k-1} (k\rho^{k-1}|z-z_1|) \right]\\
+&=|z-z_1|\sum_{n=2}^\infty|a_n|\left[\frac{n(n-1)}{2}\rho^{n-2}\right]
+\end{aligned}
+$$
+
+One can use ratio test to find that $\sum_{n=2}^\infty|a_n|\left[\frac{n(n-1)}{2}\rho^{n-2}\right]$ converges, we denote the sum using $M$
+
+So $\left|\frac{f(z)-f(z_1)}{z-z_1}-g(z_1)\right|\leq M|z-z_1|$ for $|z|<\rho$.
+
+So $\lim_{z\to z_1}\frac{f(z)-f(z_1)}{z-z_1}=g(z_1)$.
+
+QED

pages/Math416/Math416_L9.md

@@ -4,7 +4,7 @@
 
 ### Power Series
 
-Let $f(\zeta)=\sum_{n=0}^{\infty}a_n(\zeta-\zeta_0)^n$ be a power series.
+Let $f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n$ be a power series.
 
 #### Radius of Convergence
 
@@ -18,11 +18,11 @@ $$
 
 ### Derivative of Power Series
 
-Let $f(\zeta)=\sum_{n=0}^{\infty}a_n(\zeta-\zeta_0)^n$ be a power series.
+Let $f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n$ be a power series.
 
-Let $g(\zeta)=\sum_{n=0}^{\infty}na_n(\zeta-\zeta_0)^{n-1}$ be another power series.
+Let $g(z)=\sum_{n=0}^{\infty}na_n(z-z_0)^{n-1}$ be another power series.
 
-Then $g$ is holomorphic on $D(\zeta_0,R)$ and $g'(\zeta)=f(\zeta)$ for all $\zeta\in D(\zeta_0,R)$. and $f'(\zeta)=g(\zeta)$.
+Then $g$ is holomorphic on $D(z_0,R)$ and $g'(z)=f(z)$ for all $z\in D(z_0,R)$. and $f'(z)=g(z)$.
 
 Proof:
 
@@ -30,16 +30,16 @@ Note radius of convergence of $g$ is also $R$.
 
 $\limsup_{n\to\infty}|na_n|^{1/(n-1)}=\limsup_{n\to\infty}|a_n|^{1/n}$.
 
-Let $\zeta\in D(\zeta_0,R)$.
+Let $z\in D(z_0,R)$.
 
-let $|\zeta-\zeta_0|<\rho<R$.
+let $|z-z_0|<\rho<R$.
 
-Without loss of generality, assume $\zeta_0=0$. Let $|w|<\rho$.
+Without loss of generality, assume $z_0=0$. Let $|w|<\rho$.
 
 $$
 \begin{aligned}
-\frac{f(\zeta)-f(w)}{\zeta-w}-g(\zeta)&=\sum_{n=0}^{\infty}\left[\frac{1}{\zeta-w}\left(a_n(\zeta^n-w^n)\right)-na_n\zeta^{n-1}\right] \\
-&=\sum_{n=0}^{\infty}a_n\left[\frac{\zeta^n-w^n}{\zeta-w}-n\zeta^{n-1}\right]
+\frac{f(z)-f(w)}{z-w}-g(z)&=\sum_{n=0}^{\infty}\left[\frac{1}{z-w}\left(a_n(z^n-w^n)\right)-na_nz^{n-1}\right] \\
+&=\sum_{n=0}^{\infty}a_n\left[\frac{z^n-w^n}{z-w}-nz^{n-1}\right]
 \end{aligned}
 $$
 
@@ -47,24 +47,24 @@ Notice that
 
 $$
 \begin{aligned}
-\frac{\zeta^n-w^n}{\zeta-w}&=\sum_{k=0}^{n-1}\zeta^{n-1-k}w^k \\
-&=\zeta^{n-1}+\zeta^{n-2}w+\cdots+w^{n-1}
+\frac{z^n-w^n}{z-w}&=\sum_{k=0}^{n-1}z^{n-1-k}w^k \\
+&=z^{n-1}+z^{n-2}w+\cdots+w^{n-1}
 \end{aligned}
 $$
 
 Since
 
 $$
-|w^k-\zeta^k|=\left|(w-\zeta)\left(\sum_{j=0}^{k-1}w^{k-1-j}\zeta^j\right)\right|\leq|w-\zeta|k\rho^{k-1}
+|w^k-z^k|=\left|(w-z)\left(\sum_{j=0}^{k-1}w^{k-1-j}z^j\right)\right|\leq|w-z|k\rho^{k-1}
 $$
 
 $$
 \begin{aligned}
-\frac{\zeta^n-w^n}{\zeta-w}-n\zeta^{n-1}&=(\zeta^{n-1}-\zeta^{n-1})+(\zeta^{n-2}w-\zeta^{n-1})+\cdots+(\zeta w^{n-1}-\zeta^{n-1}) \\
-&=\zeta^{n-2}(w-\zeta)+\zeta^{n-3}(w^2-\zeta^2)+\cdots+\zeta^0(w^{n-1}-\zeta^{n-1}) \\
-&=\sum_{k=0}^{n-1}\zeta^{n-1-k}(w^k-\zeta^k)\\
-&\leq\sum_{k=0}^{n-1}\zeta^{n-1-k}|w-\zeta|k\rho^{k-1} \\
-&\leq|w-\zeta|\sum_{k=0}^{n-1}k\rho^{k-1} \\
+\frac{z^n-w^n}{z-w}-nz^{n-1}&=(z^{n-1}-z^{n-1})+(z^{n-2}w-z^{n-1})+\cdots+(z w^{n-1}-z^{n-1}) \\
+&=z^{n-2}(w-z)+z^{n-3}(w^2-z^2)+\cdots+z^0(w^{n-1}-z^{n-1}) \\
+&=\sum_{k=0}^{n-1}z^{n-1-k}(w^k-z^k)\\
+&\leq\sum_{k=0}^{n-1}z^{n-1-k}|w-z|k\rho^{k-1} \\
+&\leq|w-z|\sum_{k=0}^{n-1}k\rho^{k-1} \\
 \end{aligned}
 $$
 
@@ -72,13 +72,13 @@ Apply absolute value,
 
 $$
 \begin{aligned}
-\left|\frac{f(\zeta)-f(w)}{\zeta-w}-g(\zeta)\right|&\leq\sum_{n=0}^{\infty}|a_n||w-\zeta|\left[\sum_{k=1}^{n-1}\rho^{n-1-k}k\rho^{k-1}\right] \\
-&=|w-\zeta|\sum_{n=0}^{\infty}|a_n|\left[\sum_{k=1}^{n-1}\rho^{n-2}k\right] \\
-&=|w-\zeta|\sum_{n=0}^{\infty}|a_n|\frac{n(n-1)}{2}\rho^{n-2} \\
+\left|\frac{f(z)-f(w)}{z-w}-g(z)\right|&\leq\sum_{n=0}^{\infty}|a_n||w-z|\left[\sum_{k=1}^{n-1}\rho^{n-1-k}k\rho^{k-1}\right] \\
+&=|w-z|\sum_{n=0}^{\infty}|a_n|\left[\sum_{k=1}^{n-1}\rho^{n-2}k\right] \\
+&=|w-z|\sum_{n=0}^{\infty}|a_n|\frac{n(n-1)}{2}\rho^{n-2} \\
 \end{aligned}
 $$
 
-Using Cauchy-Hadamard theorem, the radius of convergence of $\sum_{n=0}^{\infty}\frac{ n(n-1)}{2}|a_n|\zeta^{n-2}$ is at least
+Using Cauchy-Hadamard theorem, the radius of convergence of $\sum_{n=0}^{\infty}\frac{ n(n-1)}{2}|a_n|z^{n-2}$ is at least
 
 $$
 1/\limsup_{n\to\infty}\left[\frac{n(n-1)}{2}|a_n|\right]^{1/(n-1)}=R.
@@ -87,63 +87,65 @@ $$
 Therefore,
 
 $$
-|w-\zeta|\sum_{n=0}^{\infty}|a_n|\frac{n(n-1)}{2}\rho^{n-2} \leq C|w-\zeta|
+|w-z|\sum_{n=0}^{\infty}|a_n|\frac{n(n-1)}{2}\rho^{n-2} \leq C|w-z|
 $$
 
 where $C$ is dependent on $\rho$.
 
-So $lim_{w\to\zeta}\left|\frac{f(\zeta)-f(w)}{\zeta-w}-g(\zeta)\right|=0$. as desired.
+So $\lim_{w\to z}\left|\frac{f(z)-f(w)}{z-w}-g(z)\right|=0$. as desired.
+
+QED
 
 #### Corollary of power series
 
-If $f(\zeta)=\sum_{n=0}^{\infty}a_n(\zeta-\zeta_0)^n$ in $D(\zeta_0,R)$, then $a_0=f(\zeta_0), a_1=f'(\zeta_0)/1!, a_2=f''(\zeta_0)/2!$, etc.
+If $f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n$ in $D(z_0,R)$, then $a_0=f(z_0), a_1=f'(z_0)/1!, a_2=f''(z_0)/2!$, etc.
 
 #### Definition (Analytic)
 
-A function $h$ on an open set $U\subset\mathbb{C}$ is called analytic if for every $\zeta\in U$, $\exists \epsilon>0$ such that on $D(\zeta,\epsilon)\subset U$, $h$ can be represented as a power series $\sum_{n=0}^{\infty}a_n(\zeta-\zeta_0)^n$.
+A function $h$ on an open set $U\subset\mathbb{C}$ is called analytic if for every $z\in U$, $\exists \epsilon>0$ such that on $D(z,\epsilon)\subset U$, $h$ can be represented as a power series $\sum_{n=0}^{\infty}a_n(z-z_0)^n$.
 
 #### Theorem (Analytic implies holomorphic)
 
 If $f$ is analytic on $U$, then $f$ is holomorphic on $U$.
 
-$\sum_{n=0}^{\infty}\frac{1}{n!}f^{(n)}(\zeta)^n$
+$\sum_{n=0}^{\infty}\frac{1}{n!}f^{(n)}(z)^n$
 
 Radius of convergence is $\infty$.
 
 So $f(0)=1=ce^0=c$
 
-$\sum_{n=0}^{\infty}\frac{1}{n}\zeta^n$
+$\sum_{n=0}^{\infty}\frac{1}{n}z^n$
 
 Radius of convergence is $1$.
 
-$f'=\sum_{n=1}^{\infty}\zeta^{n-1}=\frac{1}{1-\zeta}$ (Geometric series)
+$f'=\sum_{n=1}^{\infty}z^{n-1}=\frac{1}{1-z}$ (Geometric series)
 
-So $g(\zeta)=c+\log(\frac{1}{1-\zeta})=c+2\pi k i=\log(\frac{1}{1-\zeta})+2\pi k i$
+So $g(z)=c+\log(\frac{1}{1-z})=c+2\pi k i=\log(\frac{1}{1-z})+2\pi k i$
 
 #### Cauchy Product of power series
 
-Let $f(\zeta)=\sum_{n=0}^{\infty}a_n\zeta^n$ and $g(\zeta)=\sum_{n=0}^{\infty}b_n\zeta^n$ be two power series.
+Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$ and $g(z)=\sum_{n=0}^{\infty}b_nz^n$ be two power series.
 
-Then $f(\zeta)g(\zeta)=\sum_{n=0}^{\infty}=\sum_{n=0}^{\infty}c_n\zeta^n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_kb_{n-k}\zeta^n$
+Then $f(z)g(z)=\sum_{n=0}^{\infty}=\sum_{n=0}^{\infty}c_nz^n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_kb_{n-k}z^n$
 
 #### Theorem of radius of convergence of Cauchy product
 
-Let $f(\zeta)=\sum_{n=0}^{\infty}a_n\zeta^n$ and $g(\zeta)=\sum_{n=0}^{\infty}b_n\zeta^n$ be two power series.
+Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$ and $g(z)=\sum_{n=0}^{\infty}b_nz^n$ be two power series.
 
-Then the radius of convergence of $f(\zeta)g(\zeta)$ is at least $\min(R_f,R_g)$.
+Then the radius of convergence of $f(z)g(z)$ is at least $\min(R_f,R_g)$.
 
-Without loss of generality, assume $\zeta_0=0$.
+Without loss of generality, assume $z_0=0$.
 
 $$
 \begin{aligned}
-\left(\sum_{j=0}^{N}a_j\zeta^j\right)\left(\sum_{k=0}^{N}b_k\zeta^k\right)-\sum_{l=0}^{N}c_l\zeta^l&=\sum_{j=0}^{N}\sum_{k=N-j}^{N}a_jb_k\zeta^{j+k}\\
-&\leq\sum_{N/2\leq\max(j,k)\leq N}|a_j||b_k||\zeta^{j+k}|\\
-&\leq\left(\sum_{j=N/2}^{N}|a_j||\zeta^j|\right)\left(\sum_{k=0}^{\infty}|b_k||\zeta^k|\right)+\left(\sum_{j=0}^{\infty}|a_j||\zeta^j|\right)\left(\sum_{k=N/2}^{\infty}|b_k||\zeta^k|\right)\\
+\left(\sum_{j=0}^{N}a_jz^j\right)\left(\sum_{k=0}^{N}b_kz^k\right)-\sum_{l=0}^{N}c_lz^l&=\sum_{j=0}^{N}\sum_{k=N-j}^{N}a_jb_kz^{j+k}\\
+&\leq\sum_{N/2\leq\max(j,k)\leq N}|a_j||b_k||z^{j+k}|\\
+&\leq\left(\sum_{j=N/2}^{N}|a_j||z^j|\right)\left(\sum_{k=0}^{\infty}|b_k||z^k|\right)+\left(\sum_{j=0}^{\infty}|a_j||z^j|\right)\left(\sum_{k=N/2}^{\infty}|b_k||z^k|\right)\\
 \end{aligned}
 $$
 
-Since $\sum_{j=0}^{\infty}|a_j||\zeta^j|$ and $\sum_{k=0}^{\infty}|b_k||\zeta^k|$ are convergent, and $\sum_{j=N/2}^{N}|a_j||\zeta^j|$ and $\sum_{k=N/2}^{\infty}|b_k||\zeta^k|$ converges to zero.
+Since $\sum_{j=0}^{\infty}|a_j||z^j|$ and $\sum_{k=0}^{\infty}|b_k||z^k|$ are convergent, and $\sum_{j=N/2}^{N}|a_j||z^j|$ and $\sum_{k=N/2}^{\infty}|b_k||z^k|$ converges to zero.
 
-So $\left|\left(\sum_{j=0}^{N}a_j\zeta^j\right)\left(\sum_{k=0}^{N}b_k\zeta^k\right)-\sum_{l=0}^{N}c_l\zeta^l\right|\leq\left(\sum_{j=N/2}^{N}|a_j||\zeta^j|\right)\left(\sum_{k=0}^{\infty}|b_k||\zeta^k|\right)+\left(\sum_{j=0}^{\infty}|a_j||\zeta^j|\right)\left(\sum_{k=N/2}^{\infty}|b_k||\zeta^k|\right)\to 0$ as $N\to\infty$.
+So $\left|\left(\sum_{j=0}^{N}a_jz^j\right)\left(\sum_{k=0}^{N}b_kz^k\right)-\sum_{l=0}^{N}c_lz^l\right|\leq\left(\sum_{j=N/2}^{N}|a_j||z^j|\right)\left(\sum_{k=0}^{\infty}|b_k||z^k|\right)+\left(\sum_{j=0}^{\infty}|a_j||z^j|\right)\left(\sum_{k=N/2}^{\infty}|b_k||z^k|\right)\to 0$ as $N\to\infty$.
 
-So $\sum_{n=0}^{\infty}c_n\zeta^n$ converges to $f(\zeta)g(\zeta)$ on $D(0,R_fR_g)$.
+So $\sum_{n=0}^{\infty}c_nz^n$ converges to $f(z)g(z)$ on $D(0,R_fR_g)$.

pages/Math416/_meta.js

@@ -15,4 +15,5 @@ export default {
     Math416_L10: "Complex Variables (Lecture 10)",
     Math416_L11: "Complex Variables (Lecture 11)",
     Math416_L12: "Complex Variables (Lecture 12)",
+    Math416_L13: "Complex Variables (Lecture 13)",
 }

pages/Math416/index.md

@@ -1,3 +1,15 @@
 # Math 416
 
-Complex variables.
+Complex variables. This is a course that explores the theory and applications of complex analysis as extension of Real analysis.
+
+The course is taught by Professor.
+John E. McCarthy <mccarthy@math.wustl.edu>
+
+Some interesting fact is that he cover the lecture terribly quick. At least for me. I need to preview and review the lecture after the course ended. The only thing that I can take granted of is that many theorem in real analysis still holds in the complex. By elegant definition designing, we build a wonderful math with complex variables and extended theorems, which is more helpful when solving questions that cannot be solved in real numbers.
+
+McCarthy like to write $\zeta$ for $z$ and his writing for $\zeta$ is almost identical with $z$, I decided to use the traditional notation system I've learned to avoid confusion in my notes.
+
+I will use $B_r(z_0)$ to denote a disk in $\mathbb{C}$ such that $B_r(z_0) = \{ z \in \mathbb{C} : |z - z_0| < r \}$
+
+I will use $z$ to replace the strange notation of $\zeta$. If that makes sense.
+

pages/Swap/Math4351_L1.md

@@ -48,7 +48,7 @@ Since $r'=b(q-q')+r \geq b(q-q') \geq b$, which contradicts that $r' < b$.
 
 Therefore, $q=q'$ and $r=r'$.
 
-EOP
+QED
 
 #### Definition: Divisibility
 
@@ -74,7 +74,7 @@ Some proof examples:
 
 (2) Since $a \mid b$ and $b \mid c$, there exist $k, l \in \mathbb{Z}$ such that $b = ak$ and $c = bl$. Then $c = bl = (ak)l = a(kl)$, so $a \mid c$.
 
-EOP
+QED
 
 (3) If $a \mid b$ and $b \mid a$, then there exist $k, l \in \mathbb{Z}$ such that $b = ak$ and $a = bl$. Then $a = bl = (ak)l = a(kl)$, so $a(1-kl) = 0$.
 
@@ -82,7 +82,7 @@ Case 1: $a=0$, then $b=0$, so $a=b$.
 
 Case 2: $a \neq 0$, then $1-kl=0$, so $kl=1$. Since $k, l \in \mathbb{Z}$, $k=l=\pm 1$, so $a = \pm b$.
 
-EOP
+QED
 
 #### Definition: Divisor
 
@@ -142,7 +142,7 @@ By property of divisibility (4), $d \mid bk + (a-bk) = a$.
 
 Therefore, $d \in D(a) \cap D(b)$.
 
-EOP
+QED
 
 This theorem gives rise to the Euclidean algorithm which is a efficient way to compute the greatest common divisor of two integers. $2\Theta(\log n)+1=O(\log n)$ ([Proof in CSE 442T Lecture 7](https://notenextra.trance-0.com/CSE442T/CSE442T_L7#euclidean-algorithm)).
 

pages/index.md

@@ -10,7 +10,7 @@ So here it is. A lite server for you to read my notes.
 
 <p style="color: red; font-weight: bold">It's because I'm too easy to fall asleep if I stop doing something on my hand when my mind is wandering.</p>
 
-So I'm not responsible for any wrong notes or facts recorded in my notes since I rarely check them and use them as a reference. I'd like to read the book which have been proof read by many people.
+So I'm not responsible for any wrong notes or facts recorded in my notes since I rarely check them and use them as a reference. I'd like to read the book which have been proof read by many pQEDle.
 
 If you find any mistakes, please let me know by submitting an issue in the [GitHub repository](https://github.com/Trance-0/NoteNextra). Or click the `Edit this page` button at the side of the page to fix it yourself with appropriate push request. I really appreciate and willing to accept any valuable contribution in this project.