diff --git a/pages/CSE347/CSE347_L10.md b/pages/CSE347/CSE347_L10.md index 7436264..8b0012c 100644 --- a/pages/CSE347/CSE347_L10.md +++ b/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 diff --git a/pages/CSE347/CSE347_L5.md b/pages/CSE347/CSE347_L5.md index 9878c81..fa16035 100644 --- a/pages/CSE347/CSE347_L5.md +++ b/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) diff --git a/pages/CSE347/CSE347_L7.md b/pages/CSE347/CSE347_L7.md index c4781f7..a9ee80a 100644 --- a/pages/CSE347/CSE347_L7.md +++ b/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) diff --git a/pages/CSE347/CSE347_L8.md b/pages/CSE347/CSE347_L8.md index 9c16760..7e08ae7 100644 --- a/pages/CSE347/CSE347_L8.md +++ b/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 @@ $$ So the approximation ratio for greedy set cover is $H_d$. -EOP +QED diff --git a/pages/CSE347/CSE347_L9.md b/pages/CSE347/CSE347_L9.md index 09239f6..b9f14c0 100644 --- a/pages/CSE347/CSE347_L9.md +++ b/pages/CSE347/CSE347_L9.md @@ -296,7 +296,7 @@ If $c'\geq 8c$, then $T(n)\leq c'n\log n+1$. $E[T(n)]\leq c'n\log n+1=O(n\log n)$ -EOP +QED A more elegant proof: @@ -345,5 +345,5 @@ E[X]&=\sum_{i=0}^{n-2}\sum_{j=i+1}^{n-1}\frac{2}{j-i+1}\\ $$ -EOP +QED diff --git a/pages/CSE442T/CSE442T_L10.md b/pages/CSE442T/CSE442T_L10.md index 03417d2..8e87355 100644 --- a/pages/CSE442T/CSE442T_L10.md +++ b/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. diff --git a/pages/CSE442T/CSE442T_L12.md b/pages/CSE442T/CSE442T_L12.md index db819e7..7b334d9 100644 --- a/pages/CSE442T/CSE442T_L12.md +++ b/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 diff --git a/pages/CSE442T/CSE442T_L15.md b/pages/CSE442T/CSE442T_L15.md index bf8e856..9e57d2f 100644 --- a/pages/CSE442T/CSE442T_L15.md +++ b/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 diff --git a/pages/CSE442T/CSE442T_L16.md b/pages/CSE442T/CSE442T_L16.md index ea27f4d..6502bf9 100644 --- a/pages/CSE442T/CSE442T_L16.md +++ b/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 diff --git a/pages/CSE442T/CSE442T_L17.md b/pages/CSE442T/CSE442T_L17.md index 1a36c0c..dd6ac4c 100644 --- a/pages/CSE442T/CSE442T_L17.md +++ b/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 diff --git a/pages/CSE442T/CSE442T_L20.md b/pages/CSE442T/CSE442T_L20.md index 4b8593e..34585f9 100644 --- a/pages/CSE442T/CSE442T_L20.md +++ b/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 diff --git a/pages/CSE442T/CSE442T_L4.md b/pages/CSE442T/CSE442T_L4.md index d88564f..056690e 100644 --- a/pages/CSE442T/CSE442T_L4.md +++ b/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_ diff --git a/pages/CSE559A/CSE559A_L13.md b/pages/CSE559A/CSE559A_L13.md index 08743fe..cb0bccd 100644 --- a/pages/CSE559A/CSE559A_L13.md +++ b/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 \ No newline at end of file diff --git a/pages/CSE559A/mlp_image_reconstruction.py b/pages/CSE559A/mlp_image_reconstruction.py new file mode 100644 index 0000000..e2aa397 --- /dev/null +++ b/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() \ No newline at end of file diff --git a/pages/Math4111/Exam_reviews/Math4111_E2.md b/pages/Math4111/Exam_reviews/Math4111_E2.md index aeb0b2c..171c8a6 100644 --- a/pages/Math4111/Exam_reviews/Math4111_E2.md +++ b/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 $\frac{1}{2^n}0$ 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 \ No newline at end of file +QED \ No newline at end of file diff --git a/pages/Math4111/Math4111_L14.md b/pages/Math4111/Math4111_L14.md index 8e9ae3d..0ab16c7 100644 --- a/pages/Math4111/Math4111_L14.md +++ b/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 diff --git a/pages/Math4111/Math4111_L15.md b/pages/Math4111/Math4111_L15.md index c002e87..5805a53 100644 --- a/pages/Math4111/Math4111_L15.md +++ b/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 diff --git a/pages/Math4111/Math4111_L16.md b/pages/Math4111/Math4111_L16.md index 0629f80..62780b6 100644 --- a/pages/Math4111/Math4111_L16.md +++ b/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 diff --git a/pages/Math4111/Math4111_L17.md b/pages/Math4111/Math4111_L17.md index 497f0c6..eaa03e8 100644 --- a/pages/Math4111/Math4111_L17.md +++ b/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 diff --git a/pages/Math4111/Math4111_L18.md b/pages/Math4111/Math4111_L18.md index 06c087a..72e12e5 100644 --- a/pages/Math4111/Math4111_L18.md +++ b/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 $\lim_{n\to\infty} t_n=L$. > diff --git a/pages/Math4111/Math4111_L19.md b/pages/Math4111/Math4111_L19.md index a8c65ae..68a8f71 100644 --- a/pages/Math4111/Math4111_L19.md +++ b/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 diff --git a/pages/Math4111/Math4111_L2.md b/pages/Math4111/Math4111_L2.md index 152b519..0dc1416 100644 --- a/pages/Math4111/Math4111_L2.md +++ b/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 $E$. diff --git a/pages/Math4111/Math4111_L20.md b/pages/Math4111/Math4111_L20.md index 6a9ebfa..34dc934 100644 --- a/pages/Math4111/Math4111_L20.md +++ b/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 diff --git a/pages/Math4111/Math4111_L21.md b/pages/Math4111/Math4111_L21.md index 25f2f3e..655c47d 100644 --- a/pages/Math4111/Math4111_L21.md +++ b/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 diff --git a/pages/Math4111/Math4111_L22.md b/pages/Math4111/Math4111_L22.md index 8ee3f28..1578146 100644 --- a/pages/Math4111/Math4111_L22.md +++ b/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. diff --git a/pages/Math4111/Math4111_L24.md b/pages/Math4111/Math4111_L24.md index a50a313..d3aadbb 100644 --- a/pages/Math4111/Math4111_L24.md +++ b/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 diff --git a/pages/Math4111/Math4111_L25.md b/pages/Math4111/Math4111_L25.md index 9adf247..bf03889 100644 --- a/pages/Math4111/Math4111_L25.md +++ b/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} $$ -EOP +QED Example: diff --git a/pages/Math4111/Math4111_L3.md b/pages/Math4111/Math4111_L3.md index 39dc116..9414401 100644 --- a/pages/Math4111/Math4111_L3.md +++ b/pages/Math4111/Math4111_L3.md @@ -22,10 +22,12 @@ Let $S=\mathbb{Z}$. ## Continue -### LUBP +### LUBP (The least upper bound property) Proof that $LUBP\implies GLBP$. +Proof: + Let $S$ be an ordered set with LUBP. Let $B\alpha$ Since $(m+1)x\in \alpha$, this contradicts the fact that $\alpha$ is an upper bound of $A$. -EOP +QED ### $\mathbb{Q}$ is dense in $\mathbb{R}$ @@ -59,7 +59,7 @@ By Archimedean property, $\exist n\in \mathbb{N}$ such that $n(y-x)>1$, and $\ex So $-m_21+nx\geq 1+(m-1)=m$ -EOP +QED ### $\sqrt{2}\in \mathbb{R}$, $(\sqrt[n]{x}\in\mathbb{R})$ diff --git a/pages/Math4111/Math4111_L5.md b/pages/Math4111/Math4111_L5.md index e63d246..bb5e38c 100644 --- a/pages/Math4111/Math4111_L5.md +++ b/pages/Math4111/Math4111_L5.md @@ -94,6 +94,8 @@ So want $k\leq \frac{y^n-x}{ny^{n-1}}$ [For actual proof, see the text.] +QED + ### Complex numbers 1. $=\{a+bi:a,b\in \mathbb{R}\}$. @@ -149,7 +151,9 @@ $$ (\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. diff --git a/pages/Math4111/Math4111_L6.md b/pages/Math4111/Math4111_L6.md index ccbe1e2..0cec817 100644 --- a/pages/Math4111/Math4111_L6.md +++ b/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 diff --git a/pages/Math4111/Math4111_L7.md b/pages/Math4111/Math4111_L7.md index e40e5fa..b4a7423 100644 --- a/pages/Math4111/Math4111_L7.md +++ b/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)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)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 \ No newline at end of file +QED \ No newline at end of file diff --git a/pages/Math4111/Math4111_L9.md b/pages/Math4111/Math4111_L9.md index 9b8c466..2555979 100644 --- a/pages/Math4111/Math4111_L9.md +++ b/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 diff --git a/pages/Math4111/index.md b/pages/Math4111/index.md index 99bbf15..8ecfe39 100644 --- a/pages/Math4111/index.md +++ b/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.) diff --git a/pages/Math4121/Math4121_L10.md b/pages/Math4121/Math4121_L10.md index f34a66b..411af63 100644 --- a/pages/Math4121/Math4121_L10.md +++ b/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 diff --git a/pages/Math4121/Math4121_L11.md b/pages/Math4121/Math4121_L11.md index d5ae08f..05804e8 100644 --- a/pages/Math4121/Math4121_L11.md +++ b/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 diff --git a/pages/Math4121/Math4121_L14.md b/pages/Math4121/Math4121_L14.md index 1c4ee47..5232065 100644 --- a/pages/Math4121/Math4121_L14.md +++ b/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 diff --git a/pages/Math4121/Math4121_L15.md b/pages/Math4121/Math4121_L15.md index 394c462..c489afb 100644 --- a/pages/Math4121/Math4121_L15.md +++ b/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 diff --git a/pages/Math4121/Math4121_L16.md b/pages/Math4121/Math4121_L16.md index 3751410..fedee98 100644 --- a/pages/Math4121/Math4121_L16.md +++ b/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 diff --git a/pages/Math4121/Math4121_L17.md b/pages/Math4121/Math4121_L17.md index a68980d..fb08a93 100644 --- a/pages/Math4121/Math4121_L17.md +++ b/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 diff --git a/pages/Math4121/Math4121_L2.md b/pages/Math4121/Math4121_L2.md index ebd1f2b..33fc200 100644 --- a/pages/Math4121/Math4121_L2.md +++ b/pages/Math4121/Math4121_L2.md @@ -39,7 +39,7 @@ $$ So $h'(x)=\frac{h(t)-h(x)}{t-x}=(f'(x)+u(t))(g'(y)+v(s))$. Since $u(t)\to 0$ and $v(s)\to 0$ as $t\to x$ and $s\to y$, we have $h'(x)=g'(y)f'(x)$. -EOP +QED #### Example 5.6 @@ -135,4 +135,4 @@ If $x-A$. Apply main step, $\exists c_2\in (a,b) We take $c=\min(c_1,c_2)$. Then $\forall x\in (a,c)$, $\frac{f(x)}{g(x)} $$ -\int_{R} e^{-\zeta^2}d\zeta = 2\pi i +\int_{R} e^{-z^2}dz = 2\pi i $$ diff --git a/pages/Math416/Math416_L11.md b/pages/Math416/Math416_L11.md index 14eae39..ffde662 100644 --- a/pages/Math416/Math416_L11.md +++ b/pages/Math416/Math416_L11.md @@ -4,23 +4,23 @@ ### Continue on last example -Last lecture we have:Let $R$ be a rectangular start from the $-a$ to $a$, $a+ib$ to $-a+ib$, $\int_{R} e^{-\zeta^2}d\zeta=0$, however, the integral consists of four parts: +Last lecture we have:Let $R$ be a rectangular start from the $-a$ to $a$, $a+ib$ to $-a+ib$, $\int_{R} e^{-z^2}dz=0$, however, the integral consists of four parts: Path 1: $-a\to a$ -$\int_{I_1}e^{-\zeta^2}d\zeta=\int_{-a}^{a}e^{-\zeta^2}d\zeta=\int_{-a}^{a}e^{-x^2}dx$ +$\int_{I_1}e^{-z^2}dz=\int_{-a}^{a}e^{-z^2}dz=\int_{-a}^{a}e^{-x^2}dx$ Path 2: $a+ib\to -a+ib$ -$\int_{I_2}e^{-\zeta^2}d\zeta=\int_{a+ib}^{-a+ib}e^{-\zeta^2}d\zeta=\int_{0}^{b}e^{-(a+iy)^2}dy$ +$\int_{I_2}e^{-z^2}dz=\int_{a+ib}^{-a+ib}e^{-z^2}dz=\int_{0}^{b}e^{-(a+iy)^2}dy$ Path 3: $-a+ib\to -a-ib$ -$-\int_{I_3}e^{-\zeta^2}d\zeta=-\int_{-a+ib}^{-a-ib}e^{-\zeta^2}d\zeta=-\int_{a}^{-a}e^{-(x-ib)^2}dx$ +$-\int_{I_3}e^{-z^2}dz=-\int_{-a+ib}^{-a-ib}e^{-z^2}dz=-\int_{a}^{-a}e^{-(x-ib)^2}dx$ Path 4: $-a-ib\to a-ib$ -$-\int_{I_4}e^{-\zeta^2}d\zeta=-\int_{-a-ib}^{a-ib}e^{-\zeta^2}d\zeta=-\int_{b}^{0}e^{-(-a+iy)^2}dy$ +$-\int_{I_4}e^{-z^2}dz=-\int_{-a-ib}^{a-ib}e^{-z^2}dz=-\int_{b}^{0}e^{-(-a+iy)^2}dy$ > #### The reverse of a curve 6.9 > @@ -37,12 +37,12 @@ If we keep $b$ fixed, and let $a\to\infty$, then > > Then > -> $$\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 diff --git a/pages/Math416/Math416_L12.md b/pages/Math416/Math416_L12.md index f2780a8..0c5b0ab 100644 --- a/pages/Math416/Math416_L12.md +++ b/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 diff --git a/pages/Math416/Math416_L13.md b/pages/Math416/Math416_L13.md new file mode 100644 index 0000000..a70761e --- /dev/null +++ b/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 $|z-z_0|R$, $|\frac{z-z_0}{\zeta-z_0}|<1$. + +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 $|z|\geq R$, $\left|1+\frac{a_{n-1}}{a_nz}+\cdots+\frac{a_0}{a_nz^n}\right|<\frac{1}{2}$ \ No newline at end of file diff --git a/pages/Math416/Math416_L2.md b/pages/Math416/Math416_L2.md index 5bdc07b..482ddd2 100644 --- a/pages/Math416/Math416_L2.md +++ b/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 $$ -\frac{d}{d\zeta}\zeta^n=n\zeta^{n-1} +\frac{d}{dz}z^n=nz^{n-1} $$ diff --git a/pages/Math416/Math416_L3.md b/pages/Math416/Math416_L3.md index aad0cb6..c21beeb 100644 --- a/pages/Math416/Math416_L3.md +++ b/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} diff --git a/pages/Math416/Math416_L4.md b/pages/Math416/Math416_L4.md index 4ac9881..204dd4d 100644 --- a/pages/Math416/Math416_L4.md +++ b/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. $$ -EOP +QED If $v$ is such that $f=u+iv$ is holomorphic on $\Omega$, then $v$ is called harmonic conjugate of $u$ on $\Omega$. diff --git a/pages/Math416/Math416_L5.md b/pages/Math416/Math416_L5.md index f936403..7c06600 100644 --- a/pages/Math416/Math416_L5.md +++ b/pages/Math416/Math416_L5.md @@ -18,14 +18,14 @@ So $$ \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. diff --git a/pages/Math416/Math416_L6.md b/pages/Math416/Math416_L6.md index e5d29f0..b6e18aa 100644 --- a/pages/Math416/Math416_L6.md +++ b/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 $$ for some $k\in\mathbb{Z}$. #### Theorem 4.11 -$\log(\zeta)$ is holomorphic on $\mathbb{C}\setminus\{0\}$. +$\log(z)$ is holomorphic on $\mathbb{C}\setminus\{0\}$. Proof: Method 1: Use polar coordinates. (See in homework) -Method 2: Use the fact that $\log(\zeta)$ is the inverse of $e^\zeta$. +Method 2: Use the fact that $\log(z)$ is the inverse of $e^z$. Suppose $h=s+it$, $e^h=e^s(\cos t+i\sin t)$, $e^h-1=e^s(\cos t-1)+i\sin t$. So diff --git a/pages/Math416/Math416_L7.md b/pages/Math416/Math416_L7.md index 8b068e2..fd1b151 100644 --- a/pages/Math416/Math416_L7.md +++ b/pages/Math416/Math416_L7.md @@ -8,7 +8,7 @@ $$ 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. diff --git a/pages/Math416/Math416_L8.md b/pages/Math416/Math416_L8.md index 8ab551f..3152e95 100644 --- a/pages/Math416/Math416_L8.md +++ b/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|<\rho0$ 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 $D(0,R_fR_g)$. diff --git a/pages/Math416/_meta.js b/pages/Math416/_meta.js index 0e09d9b..d0e2643 100644 --- a/pages/Math416/_meta.js +++ b/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)", } diff --git a/pages/Math416/index.md b/pages/Math416/index.md index a3e5a6f..26ac691 100644 --- a/pages/Math416/index.md +++ b/pages/Math416/index.md @@ -1,3 +1,15 @@ # Math 416 -Complex variables. \ No newline at end of file +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 + +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. + diff --git a/pages/Swap/Math4351_L1.md b/pages/Swap/Math4351_L1.md index 9fa9945..9a37c3c 100644 --- a/pages/Swap/Math4351_L1.md +++ b/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)). diff --git a/pages/index.md b/pages/index.md index 9c06a94..01a9d23 100644 --- a/pages/index.md +++ b/pages/index.md @@ -10,7 +10,7 @@ So here it is. A lite server for you to read my notes.

It's because I'm too easy to fall asleep if I stop doing something on my hand when my mind is wandering.

-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.