Merge branch 'main' of https://github.com/Trance-0/NoteNextra

pages/Math4121/Math4121_L19.md

@@ -1 +1,44 @@
-# Lecture 19
+# Math 4121 Lecture 19
+
+## Continue on the "small set"
+
+### Cantor set
+
+#### Theorem: Cantor set is perfect, nowhere dense
+
+Proved last lecture.
+
+_Other construction of the set by removing the middle non-zero interval $(\frac{1}{n},n>0)$ and take the intersection of all such steps is called $SVC(n)$_
+
+Back to $\frac{1}{3}$ Cantor set.
+
+Every step we delete $\frac{2^{n-1}}{3^n}$ of the total "content".
+
+Thus, the total length removed after infinitely many steps is:
+
+$$
+\sum_{n=1}^{\infty} \frac{2^{n-1}}{3^n} = \frac{1}{3}\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n=1
+$$
+
+However, the quarter cantor set removes $\frac{3^{n-1}}{4^n}$ of the total "content", and the total length removed after infinitely many steps is:
+
+_skip this part, some error occurred._
+
+#### Monotonicity of outer content
+
+If $S\subseteq T$, then $c_e(S)\leq c_e(T)$.
+
+Proof: 
+
+If $C$ is cover of $T$, then $S\subseteq T\subseteq C$, so $C$ is a cover of $S$. Since $c_e(s)$ takes the inf over a larger set that $c_e(T)$, $c_e(S) \leq c_e(T)$.
+
+QED
+
+#### Theorem Osgorod's Lemma
+
+If $S$ is closed and bounded, then
+
+$$
+\lim_{k\to \infty} c_e(S_k)=c_e(S)
+$$
+

pages/Math4121/_meta.js

@@ -21,9 +21,7 @@ export default {
     Math4121_L16: "Introduction to Lebesgue Integration (Lecture 16)",
     Math4121_L17: "Introduction to Lebesgue Integration (Lecture 17)",
     Math4121_L18: "Introduction to Lebesgue Integration (Lecture 18)",
-    Math4121_L19: {
-        display: 'hidden'
-    },
+    Math4121_L19: "Introduction to Lebesgue Integration (Lecture 19)",
     Math4121_L20: {
         display: 'hidden'
     },

pages/Math416/Math416_L14.md

@@ -0,0 +1,148 @@
+# Math 416 Lecture 14
+
+## Review
+
+### Holomorphic $\iff$ Analytic
+
+#### Theorem 7.11 Liouville's Theorem
+
+Any bounded entire function is constant.
+
+## New Rollings
+
+### Finding power series for holomorphic functions
+
+Let $F$ be holomorphic on open set $U\subset \mathbb{C}$. Suppose $f(z_0)=0$, $f(z)=\sum_{n=0}^\infty a_n(z - z_0)^n$
+
+Example,
+
+$p(z)=(z-1)^3(z+i)^5(z-7)$
+
+$p(z)=\sum_{n=0}^9 c_n(z-z_0)^n$
+
+> Notice that:  
+>
+> Since $f'(z)=\sum_{n=0}^\infty a_n n(z-z_0)^{n-1}$.
+> $a_0=f(z_0)$, $a_1=f'(z_0)$, $a_k=\frac{f^{(k)}(z_0)}{k!}$ for $k \geq 0$
+
+So $c_0=0=f(1)$, $c_1=f'(1)=3(z-1)^2M=0$, $c_2=f''(1)=6(z-1)M=0$, $c_3=1$.
+
+(i) The power series for $q(z)=(z-1)^3$ at $0$.
+
+So $q(z)=\sum_{n=0}^3 a_nz^n$, you just expand it as $q(z)=z^3-3z^2+3z-1$
+
+(ii) The power series for $q(z)=(z-1)^3$ at $-1$.
+
+So $q(z)=\sum_{n=0}^3 a_n(z+1)^n$
+
+$a_0=q(-1)=(-2)^3=-8$,
+
+$a_1=q'(-1)=3(-2)^2=12$,
+
+$a_2=\frac{1}{2}q''(-1)=\frac{6}{2} \cdot -2^1 = -6$,
+
+$a_3=\frac{1}{6}q'''(-1)=\frac{6}{6} \cdot 1=1$.
+
+All higher terms are zero
+
+#### Definition: zero of multiplicity
+
+Suppose $f$ is holomorphic on open $U$ and $f(\zeta_0)=0$ for some $z_0\in U$. Let $f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$ near $z_0$. Let $m$ be the smallest number such that $a_m\neq 0$. Then we say $f$ has a zero of multiplicity $m$ at $z_0$.
+
+#### Theorem 7.12 Fundamental Theorem of Algebra
+
+Every non-constant polynomial $f$ can be factored over $\mathbb{C}$ into linear factors
+
+Proof:
+
+Since $a_n=\frac{1}{n!}f^{(n)}(z_0)$, then $f$ has a zero of order $m$ $\iff$ $f^{(m)}(z_0) \neq 0$ and $f^{(k)}(z_0) = 0, \forall k < m$.
+
+Suppose $f$ has a zero of order $m$ at $z_0$
+
+$$
+\begin{aligned}
+f(z)&=a_m(z-z_0)^m+a_{m+1}(z-z_0)^{m+1}+\cdots\\
+&=(z-z_0)^m\left[a_{m+1}(z-z_0)^{m+1}+\cdots\right]\\
+&= (z-z_0)^m g(z)
+\end{aligned}
+$$
+
+So, if $f$ has a zero of order $m$ at $\zeta_0\iff$ $f(z)=(z-z_0)^mg(z)$ where $g$ is holomorphic and $g(z_0)\neq 0$.
+
+QED
+
+#### Definition: Connected Set
+
+An open set $U$ is connected if whenever $U=U_1\cup U_2$ and $U_1, U_2$ are disjoint and open, then one of them is empty.
+
+A domain is a connected open set.
+
+#### Theorem 7.13 Zeros of Holomorphic Functions
+
+Let $U$ be a open domain (in $\mathbb{C}$). Let $f$ be holomorphic on $U$ and vanish to infinite order at some point $z_0\in U$, then $f(z)=0$ on $U$.
+
+> This is not true for $\mathbb{R}$. Consider the function $f(x) = e^{-1/x^2}$ for $x \neq 0$ and $f(0) = 0$, which is smooth and vanishes to infinite order at 0.
+
+Proof:
+
+Step 1:
+
+Show any zero of finite order is isolated.
+
+Let $z_0$ be a zero of order $m$, then by fundamental theorem of algebra, $f$ can be expressed as
+
+$$
+f(z)=(z-z_0)^mg(z)
+$$
+
+where $g$ is holomorphic and $g(z_0) \neq 0$. So $g$ is continuous.
+
+Thus $\exists$ and open set $z_0\in V$ such that $g(z_0)\neq 0$ on all of $V$.
+
+Let $U_1=\{z\in U\}$ such that $f$ vanishes to order infinity. and $U_2=U\setminus U_1$.
+
+We need to show both $U_1$ and $U_2$ are open.
+
+$U_1$:
+
+Let $z_0\in U_1$. We know that $f$ is holomorphic thus it is analytic at $z_0$.
+
+So $\exists \epsilon>0$ such that $\forall \mathbb{z}\in B_\epsilon(z_0)$
+
+So $f(z)=\sum_{n=0}^{\infty} a_n(z-z_0)^n$ implies $f(z)=0$ on $B_\epsilon(z_0)$
+
+We can expand $f$ in a power series centered at $z_1$ for any $z_1\in B_\epsilon(z_0)$, So $f(z)=\sum c_n(z-z_1)^n=0$
+
+Therefore, $z_1 \in U_1$, proving that $U_1$ is open.
+
+$U_2$:
+
+Let $w\in U_2$, if $f(w)\neq 0$, then $\exists\epsilon > 0$ such that $f(z) \neq 0$ on $B_\epsilon(w)\subset U_2$.
+
+If $f$ vanishes to finite order by Step 1, $\exists B_\epsilon(w)\subset U_2$
+
+QED
+
+#### Corollary 7.13.1 (Identity for holomorphic functions)
+
+If $f,g$ are both holomorphic on domain $U$, and they have the same power series at some point $\zeta_0$, then $f \equiv g$ on $U$.
+
+Proof:
+
+Consider $f-g$.
+
+QED
+
+#### Corollary 7.13.2
+
+Let $U$ be a domain, $f\in O(U)$, $f$ is not identically zero on $U$, $f^{-1}(0)$ has no limit point on $U$.
+
+Proof:
+
+We proceed by contradiction. Suppose $z_n\to w\in U$, $f(z_0)=0$, $f(w)=0$. $w$ is not an isolated zero. So $f$ is a zero of infinite order. Contradicting with our assumption that $f$ is not identically zero.
+
+QED
+
+#### Corollary 7.13.3: Identity principle
+
+If $f,g\in O(U)$, $U$ is a domain and $\exists$ sequence $z_0$ that converges to $w\in U$, such that $f(z_n)=g(z_n)$, then $f\equiv g$ on U$.

pages/Math416/_meta.js

@@ -16,4 +16,5 @@ export default {
     Math416_L11: "Complex Variables (Lecture 11)",
     Math416_L12: "Complex Variables (Lecture 12)",
     Math416_L13: "Complex Variables (Lecture 13)",
+    Math416_L14: "Complex Variables (Lecture 14)",
 }