update

pages/Math4121/Math4121_L3.md

@@ -1 +1,86 @@
-# Lecture 3
+# Lecture 3
+
+## Continue on Differentiation
+
+### Mean Value Theorem
+
+#### Theorem 5.9 Generalized Mean Value Theorem
+
+If $f,g:[a,b]\to \mathbb{R}$ are continuous on $[a,b]$ and differentiable in $(a,b)$, then there exists a point $x\in (a,b)$ such that
+
+$$
+[f(b)-f(a)]g'(x)=[g(b)-g(a)]f'(x)
+$$
+
+Proof:
+
+Define $h(x)=[f(b)-f(a)]g(x)-[g(b)-g(a)]f(x)$, $t\in [a,b]$.
+
+We need to show that there exists a point $x\in (a,b)$ such that $h'(x)=0$.
+
+By previous theorem, it's enough to show that $h$ has a local minimum or maximum in $(a,b)$. By [Extreme Value Theorem](https://notenextra.trance-0.com/Math4111/Math4111_L24#theorem-416-extreme-value-theorem)
+
+$$
+\begin{aligned}
+h(a)&=[f(b)-f(a)]g(a)-[g(b)-g(a)]f(a)\\
+&=f(b)g(a)-f(a)g(b)\\
+h(b)&=[f(b)-f(a)]g(b)-[g(b)-g(a)]f(b)\\
+&=g(a)f(b)-g(b)f(a)
+\end{aligned}
+$$
+
+So $h(a)=h(b)$.
+
+Consider three cases:
+
+1. $h$ is constant on $[a,b]$. Then $h'(x)=0$ for all $x\in (a,b)$.
+2. $\exists t\in (a,b)$ such that $h(t)>h(a)=h(b)$. Since every continuous function on a compact interval attains its supremum, and $h(t)>h(a)=h(b)$, the supremum of $h$ on $[a,b]$ is attained at some point $x\in (a,b)$. (Apply [Extreme Value Theorem](https://notenextra.trance-0.com/Math4111/Math4111_L24#theorem-416-extreme-value-theorem) to $h$ on $[a,b]$.)
+3. $\exists t\in (a,b)$ such that $h(t)<h(a)=h(b)$. Since every continuous function on a compact interval attains its infimum, and $h(t)<h(a)=h(b)$, the infimum of $h$ on $[a,b]$ is attained at some point $x\in (a,b)$. (Apply [Extreme Value Theorem](https://notenextra.trance-0.com/Math4111/Math4111_L24#theorem-416-extreme-value-theorem) to $h$ on $[a,b]$.)
+
+In all cases, $h$ has a local minimum or maximum in $(a,b)$.
+
+EOP
+
+#### Theorem 5.10 Mean Value Theorem
+
+The mean value theorem is a special case of the generalized mean value theorem when $g(x)=x$ (the identity function).
+
+If $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a point $x\in (a,b)$ such that
+
+$$
+f(b)-f(a)=f'(x)(b-a)
+$$
+
+### Intermediate Value Theorem
+
+#### Definition 5.12.1 Intermediate Value
+
+We say that $f:[a,b]\to \mathbb{R}$ attains the intermediate values if for each $\lambda\in (f(a),f(b))$ there exists a point $x\in (a,b)$ such that $f(x)=\lambda$.
+
+#### Theorem 5.12.2 Continuous Function attains Intermediate Values
+
+If $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$, then $f$ attains every value between $f(a)$ and $f(b)$.
+
+#### Theorem 5.12 Intermediate Value Theorem
+
+If $f:[a,b]\to \mathbb{R}$ is differentiable on $[a,b]$. Then $f'$ attains intermediate values.
+
+Proof:
+
+Let $\lambda\in (f'(a),f'(b))$.
+
+Let our auxiliary function be $g(t)=f(t)-\lambda t$.
+
+Since $g'(t)=f'(t)-\lambda$, it suffices to find $x\in (a,b)$ such that $g'(x)=0$.
+
+$g'(a)<0$ and $g'(b)>0$.
+
+We claim that $\exists t_1\in (a,b)$ such that $g(t_1)<g(a)$.
+
+If this were false, then for all $t\in (a,b)$ we would have $g(t)\geq g(a)$.
+
+$$
+\frac{g(t)-g(a)}{t-a}\geq 0\\
+$$
+
+Continue on Monday.

pages/Math4121/_meta.js

@@ -5,12 +5,8 @@ export default {
     },
     Math4121_L1: "Introduction to Lebesgue Integration (Lecture 1)",
     Math4121_L2: "Introduction to Lebesgue Integration (Lecture 2)",
-    Math4121_L3: {
-        display: 'hidden'
-    },
-    Math4121_L4: {
-        display: 'hidden'
-    },
+    Math4121_L3: "Introduction to Lebesgue Integration (Lecture 3)",
+    Math4121_L4: "Introduction to Lebesgue Integration (Lecture 4)",
     Math4121_L5: {
         display: 'hidden'
     },