From f46e9de9cbb29092dd199f3dc6aef711e76a90e7 Mon Sep 17 00:00:00 2001 From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com> Date: Sun, 2 Mar 2025 10:50:06 -0600 Subject: [PATCH] update --- pages/Math4121/Math4121_L19.md | 45 +++++++++- pages/Math4121/_meta.js | 4 +- pages/Math416/Math416_L14.md | 148 +++++++++++++++++++++++++++++++++ pages/Math416/_meta.js | 1 + 4 files changed, 194 insertions(+), 4 deletions(-) create mode 100644 pages/Math416/Math416_L14.md diff --git a/pages/Math4121/Math4121_L19.md b/pages/Math4121/Math4121_L19.md index ca72bc8..a9a5f2d 100644 --- a/pages/Math4121/Math4121_L19.md +++ b/pages/Math4121/Math4121_L19.md @@ -1 +1,44 @@ -# Lecture 19 \ No newline at end of file +# 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 + +$$ +\lim_{k\to \infty} c_e(S_k)=c_e(S) +$$ + diff --git a/pages/Math4121/_meta.js b/pages/Math4121/_meta.js index 3586b26..04a934f 100644 --- a/pages/Math4121/_meta.js +++ b/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' }, diff --git a/pages/Math416/Math416_L14.md b/pages/Math416/Math416_L14.md new file mode 100644 index 0000000..1f0186e --- /dev/null +++ b/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$. \ No newline at end of file diff --git a/pages/Math416/_meta.js b/pages/Math416/_meta.js index d0e2643..2d5e369 100644 --- a/pages/Math416/_meta.js +++ b/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)", }