update

next.config.mjs

@@ -5,15 +5,20 @@ const withNextra = nextra({
   themeConfig: './theme.config.jsx',
   latex: {
     renderer: 'katex',
-    // options: {
-    //   macros: {
-    //     '\\RR': '\\mathbb{R}'
-    //   }
-    // }
+    options: {
+      // suppress warnings from katex for `\\`
+      strict: false,
+      // macros: {
+      //   '\\RR': '\\mathbb{R}'
+      // }
+    }
   }
 })
 
 export default withNextra({
+  eslint: {
+    ignoreDuringBuilds: true,
+  },
   experimental: {
     // optimize memory usage: https://nextjs.org/docs/app/building-your-application/optimizing/memory-usage
     webpackMemoryOptimizations: true,

pages/Math416/Math416_L11.md

@@ -109,7 +109,7 @@ Cauchy's theorem is true if $\gamma$ is a triangle.
 
 Proof:
 
-We plan to keep shrinking the triangle until $f(\zeta+h)=f(\zeta)+hf'(\zeta)+\mu(h)$ where $\mu(h)$ is a function of $h$ that goes to $0$ as $h\to 0$.
+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 $h\to 0$.
 
 Let's start with a triangle $T$ with vertices $z_1,z_2,z_3$.
 
@@ -117,5 +117,53 @@ Let's start with a triangle $T$ with vertices $z_1,z_2,z_3$.
 
 We divide $T$ into four smaller triangles by drawing lines from the midpoints of the sides to the opposite vertices.
 
+Let $R_1,\ldots,R_4$ be the four smaller triangles.
 
+For one $R_j$, $\left|\int_{R_j}f(z)dz\right|\geq\frac{1}{4}|I|$, we choose it then call it $T_1$.
 
+There exists $T_1$ such that $\left|\int_{T_1}f(z)dz\right|\geq\frac{1}{4}|I|$.
+
+Since $L(T_1)=\frac{1}{2}L(T)$, we iterate after $n$ steps, get a triangle $T_n$ such that $L(T_n)=\frac{L(T)}{2^n}$ and $\left|\int_{T_n}f(z)dz\right|\geq\frac{1}{4^n}|I|$.
+
+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.
+
+So $f(\zeta)=f(z_0)+f'(z_0)(\zeta-z_0)+R(\zeta)$, we have
+
+$$
+\int_{T_n}f(\zeta)d\zeta=\int_{T_n}f(z_0)d\zeta+\int_{T_n}f'(z_0)(\zeta-z_0)d\zeta+\int_{T_n}R(\zeta)d\zeta
+$$
+
+since $f(z_0)d\zeta+\int_{T_n}f'(z_0)(\zeta-z_0)$ is in form of Cauchy integral formula, we have
+
+$$
+\int_{T_n}f(z_0)d\zeta+\int_{T_n}f'(z_0)(\zeta-z_0)d\zeta=0
+$$
+
+Let $e_n=\max\{\frac{R(\zeta)}{\zeta-z_0}:z_0\in T_n\}$
+
+Since $diam(K_n)\to 0$ as $n\to\infty$, we have $e_n\to 0$ as $n\to\infty$.
+
+So
+
+$$
+\begin{aligned}
+|I|&\leq 4^n\left|\int_{T_n}f(\zeta)d\zeta\right|\\
+&\leq 4^n\left|\int_{T_n}R_n(\zeta)d\zeta\right|\\
+&\leq 4^n\cdot L(T_n)\cdot \max_{\zeta\in T_n}|R_n(\zeta)|\\
+&\leq 4^n\cdot \frac{L(T_0)}{2^n}\cdot e_n L(T_n)\\
+&\leq 4^n\cdot \frac{L(T_0)}{2^n}\cdot e_n\cdot \frac{L(T_0)}{2^n}\\
+&\leq e_n\cdot L(T_0)^2
+\end{aligned}
+$$
+
+Since $e_n\to 0$ as $n\to\infty$, we have $I\to 0$ as $n\to\infty$.
+
+So
+
+$$
+\int_{T_n}f(\zeta)d\zeta\to 0
+$$
+
+EOP