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

pages/Math4121/Math4121_L2.md

@@ -1 +1,138 @@
-# Lecture 2
+# Lecture 2
+
+## Chapter 5: Differentiation
+
+### Continue on Differentiation
+
+#### Theorem 5.5: Chain Rule
+
+Suppose
+
+1. $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ (or some neighborhood of $x$)
+2. $f'(x)$ exists at some point $x\in (a,b)$ ($f$ is differentiable at $x$)
+3. $g$ is defined on an interval $[c,d]$ containing the range of $f$, ($f([a,b])\subset [c,d]$)
+4. $g$ is differentiable at the point $f(x)$
+
+Let $h=g\circ f$ ($h=g(f(x))$) where $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$. Then $h$ is differentiable at $x$ and
+
+$$
+h'(x) = g'(f(x))f'(x)
+$$
+
+Proof:
+
+Let $y=f(x)$ and $u(t)=\frac{f(t)-f(x)}{t-x}-f'(x)$ for $t\neq x,t\in [a,b]$, $v(s)=\frac{g(s)-g(y)}{s-y}-g'(y)$ for $s\neq y,s\in [c,d]$.
+
+Notice that $u(t)\to 0$ as $t\to x$ and $v(s)\to 0$ as $s\to y$.
+
+Pick $s=f(t)$ for $t\in [a,b]$ so that $s\to y$ as $t\to x$. Then
+
+$$
+\begin{aligned}
+h(t)-h(x) &= g(f(t))-g(f(x)) \\
+&= g(t)-g(y) \\
+&= (s-y)(g'(y)+v(s)) \\
+&= (f(t)-f(x))(g'(y)+v(s)) \\
+&= (t-x)(f'(x)+u(t))(g'(y)+v(s)) \\
+\end{aligned}
+$$
+
+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
+
+#### Example 5.6
+
+(a) Let $f(x)=\begin{cases}
+x\sin\frac{1}{x} & x\neq 0 \\
+0 & x=0
+\end{cases}$
+
+For $x\neq 0$,
+
+$$
+\begin{aligned}
+f'(x) &= 1\cdot\sin\frac{1}{x}+x\cos\frac{1}{x}\cdot\frac{-1}{x^2} \\
+&= \sin\frac{1}{x}-\frac{\cos\frac{1}{x}}{x}
+\end{aligned}
+$$
+
+For $x=0$,
+
+$$
+\begin{aligned}
+f'(0) &= \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} \\
+&= \lim_{x\to 0}\frac{x\sin\frac{1}{x}}{x} \\
+&= \lim_{x\to 0}\sin\frac{1}{x}
+\end{aligned}
+$$
+
+This limit does not exist, so $f$ is not differentiable at $x=0$.
+
+(b) Let $f(x)=\begin{cases}
+x^2 \sin\frac{1}{x} & x\neq 0 \\
+0 & x=0
+\end{cases}$
+
+For $x\neq 0$,
+
+$$
+\begin{aligned}
+f'(x) &= 2x\sin\frac{1}{x}+x^2\cos\frac{1}{x}\cdot\frac{-1}{x^2} \\
+&= 2x\sin\frac{1}{x}-\cos\frac{1}{x}
+\end{aligned}
+$$
+
+For $x=0$,
+
+$$
+\begin{aligned}
+f'(0) &= \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} \\
+&= \lim_{x\to 0}\frac{x^2\sin\frac{1}{x}}{x} \\
+&= \lim_{x\to 0}x\sin\frac{1}{x}\\
+&= 0
+\end{aligned}
+$$
+
+So $f'(x)=\begin{cases}
+2x\sin\frac{1}{x}-\cos\frac{1}{x} & x\neq 0 \\
+0 & x=0
+\end{cases}$.
+
+Notice that $f'(x)$ is not continuous at $x=0$ since $\lim_{x\to 0}f'(x)$ is undefined.
+
+### Mean Value Theorem
+
+#### Definition 5.7: Local Extrema
+
+Let $f:[a,b]\to \mathbb{R}$. We say that $f$ has a local maximum (or minimum) at $x\in [a,b]$ if there exists some $\delta>0$ such that 
+
+$$
+f(x)\geq f(t) \text{ for all }|x-t|<\delta
+$$
+
+for local maximum, and
+
+$$
+f(x)\leq f(t) \text{ for all }|x-t|<\delta
+$$
+
+for local minimum.
+
+#### Theorem 5.8
+
+If $f:[a,b]\to \mathbb{R}$ has a local maximum (or minimum) at $x\in (a,b)$ and $f$ is differentiable at $x$, then $f'(x)=0$.
+
+Proof:
+
+We can find $\delta>0$ such that $a<x-\delta<x<x+\delta<b$.
+
+And for all $x-\delta<t<x+\delta$,
+
+If $x-\delta<t<x$, then $f(x)\geq f(t)$ so $\frac{f(t)-f(x)}{t-x}\leq 0$.
+
+If $x<t<x+\delta$, then $f(x)\geq f(t)$ so $\frac{f(t)-f(x)}{t-x}\geq 0$.
+
+So $\lim_{t\to x}\frac{f(t)-f(x)}{t-x}=0$.
+
+EOP

pages/Math4121/_meta.js

@@ -4,9 +4,7 @@ export default {
         type: 'separator'
     },
     Math4121_L1: "Introduction to Lebesgue Integration (Lecture 1)",
-    Math4121_L2: {
-        display: 'hidden'
-    },
+    Math4121_L2: "Introduction to Lebesgue Integration (Lecture 2)",
     Math4121_L3: {
         display: 'hidden'
     },

pages/Math4121/index.md

@@ -21,3 +21,10 @@ A radical Approach to Lebesgue's Theory of Integration by David
 
 Due every Monday.
 
+## Office Hour
+
+Monday 1pm-2pm
+
+Wednesday 12pm-1pm
+
+Friday 11:30am-12:30pm