update

docker-compose.yaml

@@ -3,7 +3,7 @@ services:
     build:
       context: ./
       dockerfile: ./Dockerfile
-    image: trance0/notenextra:v1.1.9
+    image: trance0/notenextra:v1.1.10
     restart: on-failure:5
     ports:
       - 13000:3000

pages/Math4121/_meta.js

@@ -27,15 +27,9 @@ export default {
     Math4121_L22: "Introduction to Lebesgue Integration (Lecture 22)",
     Math4121_L23: "Introduction to Lebesgue Integration (Lecture 23)",
     Math4121_L24: "Introduction to Lebesgue Integration (Lecture 24)",
-    Math4121_L25: {
+    Math4121_L25: "Introduction to Lebesgue Integration (Lecture 25)",
-        display: 'hidden'
+    Math4121_L26: "Introduction to Lebesgue Integration (Lecture 26)",
-    },
+    Math4121_L27: "Introduction to Lebesgue Integration (Lecture 27)",
-    Math4121_L26: {
-        display: 'hidden'
-    },
-    Math4121_L27: {
-        display: 'hidden'
-    },
     Math4121_L28: {
         display: 'hidden'
     },

pages/Math416/Math416_L18.md

@@ -0,0 +1,188 @@
+# Math416 Lecture 18
+
+## Chapter 8: Laurent Series and Isolated Singularities
+
+### 8.1 Laurent Series
+
+#### Definition of Laurent Series
+
+$$
+\sum_{n=-\infty}^{\infty} c_n (z-z_0)^n
+$$
+
+where $c_n$ are complex coefficients.
+
+Let $R_2=\frac{1}{\limsup_{n\to\infty} |c_n|^{1/n}}$, then the Laurent series converges on $|z-z_0|<R_2$
+
+Where $R_1=\limsup_{n\to-\infty} |c_n|^{-1/n}$, if $|z-z_0|>R_1$, the Laurent series diverges.
+
+If $R_1\leq R_2$, then the Laurent series converges on $A(z_0;R_1,R_2)=\{z:R_1<|z-z_0|<R_2\}$, the Laurent series converges absolutely on $A(z_0;R_1,R_2)$
+
+By Weierstrass, the limit is a holomorphic function on $A(z_0;R_1,R_2)$
+
+If $R_1<r<R_2$, then
+
+$$
+\int_{C(z_0,r)} f(z) dz = \sum_{n=-\infty}^{\infty} c_n \int_{C(z_0,r)} (z-z_0)^n dz
+$$
+
+> $$
+\int_{C(z_0,r)} (z-z_0)^n dz = \begin{cases}
+    2\pi i, & n=-1 \\
+    0, & n\neq -1
+\end{cases}$$
+> Proof:
+> $\gamma(t)=z_0+re^{it}, t\in[0,2\pi]$
+> $$\begin{aligned}
+\int_{C(z_0,r)} (z-z_0)^n dz &= \int_0^{2\pi} (z_0+re^{it}-z_0)^n ire^{it} dt \\
+&= ir^{n+1} \int_0^{2\pi} e^{i(n+1)t} dt \\
+&= \begin{cases}
+    2\pi i, & n=-1 \\
+    \int_0^{2\pi} e^{i(n+1)t} dt = \frac{1}{i(n+1)}e^{i(n+1)t}\Big|_0^{2\pi} = 0, & n\neq -1
+\end{cases}
+\end{aligned}$$
+
+So,
+
+$$
+\int_{C(z_0,r)} f(z) dz = \sum_{n=-\infty}^{\infty} c_n \int_{C(z_0,r)} (z-z_0)^n dz=c_{-1}2\pi i
+$$
+
+And,
+
+$$
+\int_{C(z_0,r)} f(z)(z-z_0)^k dz = \sum_{n=-\infty}^{\infty} c_n \int_{C(z_0,r)} (z-z_0)^{n+k} dz = c_{-1-k}2\pi i
+$$
+
+So,
+
+$$
+2\pi i c_j = \int_{C(z_0,r)} f(z)(z-z_0)^{-j-1} dz
+$$
+
+### Cauchy integral
+
+Recall Cauchy integral formula:
+
+$$
+f(z) = \int_{\gamma} \frac{\phi(\xi)}{\xi-z} d\xi
+$$
+
+where $\gamma$ is a closed curve.
+
+Suppose $|z-z_0|>R$,
+
+$$
+\begin{aligned}
+\frac{1}{\xi-z}&=\frac{1}{\xi-z_0-(z-z_0)}\\
+&=-\frac{1}{z-z_0}\frac{1}{1-\frac{\xi-z_0}{z-z_0}}\\
+&=-\frac{1}{z-z_0}\sum_{n=0}^{\infty} \frac{(\xi-z_0)^n}{(z-z_0)^n}\\
+&=-\sum_{m=1}^{\infty} (\xi-z_0)^{m-1}(z-z_0)^{-m}
+\end{aligned}
+$$
+
+So,
+
+$$
+\begin{aligned}
+f(z) &= \int_{\gamma} \frac{\phi(\xi)}{\xi-z} d\xi\\
+&= -\int_{\gamma} \sum_{m=1}^{\infty} (\xi-z_0)^{m-1}(z-z_0)^{-m}\phi(\xi) d\xi\\
+&=-\sum_{m=1}^{\infty} (z-z_0)^{-m} \int_{\gamma} (\xi-z_0)^{m-1} \phi(\xi) d\xi
+\end{aligned}
+$$
+
+So the Cauchy integral $\int_{\gamma} \frac{\phi(\xi)}{\xi-z} d\xi$ is a convergent power series in $B_{d(z_0,\gamma)}(z_0)$ and is a convergent Laurent series (with just negative powers) in $\mathbb{C}\setminus B_{\max_{\xi\in \gamma} d(z_0,\xi)}(z_0)$
+
+#### Theorem 8.4 Cauchy Theorem for Annulus
+
+Suppose $f$ is holomorphic on $A(z_0;R_1,R_2)$, Let $C_r=\{z:|z-z_0|=r\}$, oriented counterclockwise. Then $I(r)=\int_{C_r} f(z) dz$ is independent of $r$ for $R_1<r<R_2$
+
+Proof:
+
+If integrand is continuous with respect to $r$ and continuous with respect to $t$, then we can differentiate under the integral sign (Check after class, on Appendix 4?)
+
+$$
+\begin{aligned}
+I(r)&=\int_0^{2\pi} f(z_0+re^{it})ire^{it}dt\\
+\frac{dI}{dr}&=i\int_0^{2\pi}\frac{\partial}{\partial r}[f(z_0+re^{it})re^{it}]dt
+\end{aligned}
+$$
+
+$$
+\frac{\partial}{\partial r}[f(z_0+re^{it})ire^{it}]=if'(z_0+re^{it})e^{it}+if(z_0+re^{it})e^{it}
+$$
+
+$$
+\frac{\partial}{\partial t}[f(z_0+re^{it})e^{it}]=f'(z_0+re^{it})ire^{it}+f(z_0+re^{it})ire^{it}
+$$
+
+This gives
+
+$$
+\frac{\partial}{\partial r}[f(z_0+re^{it})ire^{it}]=\frac{\partial}{\partial t}[f(z_0+re^{it})e^{it}]
+$$
+
+So,
+
+$$
+\frac{dI}{dr}=i\int_0^{2\pi}\frac{\partial}{\partial r}[f(z_0+re^{it})ire^{it}]dt=i\int_0^{2\pi}\frac{\partial}{\partial t}[f(z_0+re^{it})e^{it}]dt=0
+$$
+
+is a integration on a closed curve, so it is $0$.
+
+So, $I(r)$ is constant.
+
+QED
+
+Let $f$ be holomorphic on $A(z_0;R_1,R_2)$. Let $C_r=\{z:|z-z_0|=r\}$, oriented counterclockwise. Let $w\in A(z_0;R_1,R_2)$. Choose $R_1<r_1<|w-z_0|<r_2<R_2$ such that $w\in A(z_0;r_1,r_2)$. Then,
+
+$$
+f(w)=\frac{1}{2\pi i}\int_{C_{r_2}} \frac{f(z)}{z-w} dz-\frac{1}{2\pi i}\int_{C_{r_1}} \frac{f(z)}{z-w} dz
+$$
+
+Proof:
+
+Define $g(z)=\begin{cases}
+    \frac{f(z)-f(w)}{z-w}, & z\neq w \\
+    f'(w), & z=w
+\end{cases}$
+
+Then $g$ is holomorphic on $A(z_0;r_1,r_2)$ since $f$ is analytic at $w$, $f(z)=\sum_{n=0}^{\infty} \frac{f^{(n)}(w)}{n!}(z-w)^n$
+
+So,
+
+$$
+f(z)-f(w)=f(w)+f'(w)(z-w)+\sum_{n=2}^{\infty} \frac{f^{(n)}(w)}{n!}(z-w)^n-f(w)=f'(w)(z-w)+\sum_{n=2}^{\infty} \frac{f^{(n)}(w)}{n!}(z-w)^n
+$$
+
+So,
+
+$$
+\frac{f(z)-f(w)}{z-w}=f'(w)+\sum_{n=1}^{\infty} \frac{f^{(n)}(w)}{n!}(z-w)^{n-1}
+$$
+
+So,
+
+$$
+\lim_{z\to w} \frac{f(z)-f(w)}{z-w}=f'(w)
+$$
+
+So $\int_{C_{r_2}} g(z) dz=\int_{C_{r_1}} g(z) dz$,
+
+$$
+\int_{C_{r_2}} \frac{f(z)}{z-w} dz-f(w)\int_{C_{r_2}} \frac{1}{z-w} dz=\int_{C_{r_1}} \frac{f(z)}{z-w} dz-f(w)\int_{C_{r_1}} \frac{1}{z-w} dz
+$$
+
+Since $\int_{C_{r_2}} \frac{1}{z-w} dz=2\pi i$ and $\int_{C_{r_1}} \frac{1}{z-w} dz=0$, (using Cauchy integral theorem on convex region)
+
+$$
+f(w)=\frac{1}{2\pi i}\int_{C_{r_2}} \frac{f(z)}{z-w} dz-\frac{1}{2\pi i}\int_{C_{r_1}} \frac{f(z)}{z-w} dz
+$$
+
+QED
+
+Since $\int_{C_{r_1}} \frac{f(z)}{z-w} dz$ is a Laurent series in negative powers which converges in $\mathbb{C}\setminus \overline{B_{r_1}(z_0)}$, we can conclude that
+
+$f(z)$ is given by a convergent Laurent series $\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$ in $\mathbb{C}\setminus \overline{B_{r_1}(z_0)}$ where $a_n=\frac{1}{2\pi i}\int_{C_r} \frac{f(z)}{(z-z_0)^{-1-n}} dz$
+
+Laurent series converges in $A(z_0;R_1,R_2)$

pages/Math416/_meta.js

@@ -21,4 +21,5 @@ export default {
     Math416_L15: "Complex Variables (Lecture 15)",
     Math416_L16: "Complex Variables (Lecture 16)",
     Math416_L17: "Complex Variables (Lecture 17)",
+    Math416_L18: "Complex Variables (Lecture 18)",
 }

pages/contact.md

@@ -6,5 +6,5 @@ This page is mainly maintained by [Trance-0](https://github.com/Trance-0).
 
 [GitHub](https://github.com/Trance-0)
 
-[Email](mailto:wuzheyuan.86@gmail.com)
+[Email](mailto:me@trance-0.com)