182 lines
4.2 KiB
Markdown
182 lines
4.2 KiB
Markdown
# Lecture 5
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## Review
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In each case, determine (with justification) whether the claim or its negation is the true statement.
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(a) For all real numbers satisfying $a<b$, there exists an $n\in \mathbb{N}$ such that $a+1/n<b$.
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negation: $\exists a<b$, $\forall n\in \mathbb{N}$ such that $a+1/n\geq b$.
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By Archimedean property, the statement is true.
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(b) There exists a real number $x>0$ such that $x<1/n$ for all $n\in \mathbb{N}$.
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The statement is ambiguous because we can arrange the statement in two ways.
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$\exists x\in \mathbb{R}_{>0}$ such that $\forall n\in \mathbb{N},x\leq \frac{1}{n}$
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negation: $\forall x\in \mathbb{R}_{>0}$, $\exists n\in \mathbb{N}$, such that $x\leq \frac{1}{n}$.
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The statement is true, let $x=\frac{1}{n+1}$.
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## New Materials
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### Continue on the theorem
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#### Theorem 1.21
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$\forall x\in \mathbb{R}_{>0},\forall n\in \mathbb{N},\exist$ unique $y\in \mathbb{R}_{>0}$ such that $y^n=x$.
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(Because of this Theorem we can define $x^{1/x}=y$ and $\sqrt{x}=y$)
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<details>
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<summary>Proof</summary>
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We cna assume $n\geq 2$ (For $n=1,y=x$)
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Step 1 (uniqueness): If $0<y_1<y_2$, then $y_1^n<y_2^n$ (by properties of ordered field)
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Step 2 (existence): Let $E=\{t\in \mathbb{R}_{>0}: t^n<x\}$ We want to let $y=sup\ E$, but to do this we need to check 2 things.
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1. To show $E\neq \phi$:
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If $x\geq 1$, then $1/2\in E$.
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If $x<1$, then $x\in E$.
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2. To show $E$ is bounded above. We need to find an upper bound.
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If $x\geq 1$, then $x\in E$
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If $x<1$, then $1 \in E$.
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So we can let $y=sup\ E$
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Step 2b ($y^n\geq x$) Suppose for contradiction $y^n<x$.
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Thoughts: If we can find $h>0$ such that $(y+h)^n<x$, then $y+h\in E$. This would contradict the facts $y$ is an upper bound of $E$.
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$$
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(y+h)^n=y^n+ny^{n-1}h+{more\ terms}
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$$
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We want $ny^{n-1}h+{more\ terms}<x-y^n$
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Observe: If $0<a<b$, then
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$$
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\frac{b^n-a^n}{b-a}=b^{n-1}+b^{n-2}a+...+a^{n-1}\leq b^{n-1}+b^{n-2}b+...+b^{n-1}=nb^{n-1}
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$$
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The fact tells us $(y+h)^n-y^n\leq n(y+h)^{n-1}h$.
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And we want $n(y+h)^{n-1} h+y^n<x$.
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So want $h<\frac{x-y^h}{n(y+h)^{n-1}}$.
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To achieve this, choosey any $h>0$ satisfying $h<1$ and $h<\frac{x-y^h}{n(y+h)^{n-1}}$
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[For actual proof, see the text.]
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Step 2c showing ($y^n\leq x$)
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Suppose for contradiction $y^n>x$
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Thoughts: Find $k>0$ such that $(y-k)^n>x$.
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Then $y-k$ is an upper bound for $E$, which contradicts the fact that $y$ is the least upper bound of $E$.
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$y^n-(y-k)^n\leq ny^{n-1}k$.
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We want $y^n-ny^{n-1}k\geq x$.
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So want $k\leq \frac{y^n-x}{ny^{n-1}}$
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[For actual proof, see the text.]
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</details>
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### Complex numbers
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1. $=\{a+bi:a,b\in \mathbb{R}\}$.
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Conjugate: $z=a+bi,\bar{z}=a-bi$.
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#### Theorem 1.31 (see text)
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Pure computational proof: boring...
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$z\bar{z}=a^2-(bi)^2=a^2+b^2$
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You can also use vector sum for representing operation in complex numbers.
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#### Theorem 1.33 (see text)
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More computation and still, boring...
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some fun theorems:
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- $|Re\ z|\leq |z|$ (equal when no imaginary parts)
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- $|z+w|\leq |z|+|w|$ (equal when both $z,w$ have no imaginary parts) (Triangle inequality)
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Proof for $|z+w|\leq |z|+|w|$:
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$$
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|z+w|^2=(z+w)(\overline{z+w})=(z+w)(\bar{z}+\bar{w})=|z|^2+|w|^2+z\bar{w}+\bar{z}w
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$$
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Since
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$$
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z\bar{w}+\bar{z}w\leq 2Re(z\bar{w})
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$$
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$$
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(z+w)(\bar{z}+\bar{w})=|z|^2+|w|^2+z\bar{w}+\bar{z}w\leq |z|^2+|w|^2+2|z||w|\leq |z|+|w|
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$$
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#### Theorem 1.35 Cauchy-Schwarz inequality
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If $\vec{a},\vec{b}\in \mathbb{C}^n$, then
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$$
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|\vec{a}\vec{b}|^2\leq (\vec{a}\vec{a})(\vec{b}\vec{b})
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$$
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> Remark: The proof is very tricky.
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To help us motivate the proof in text, let's consider the special case of real numbers.
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$$
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(\sum a_j b_j)^2=(\sum a_j^2)(\sum b_j^2)
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$$
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<details>
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<summary>Proof</summary>
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For real numbers:
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Let $A=\sum a_j^2,B=\sum b_j^2, C=\sum a_j b_j$, want to show $C^2\leq AB$
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Note: if $B=0$, then $b_1=b_2=...=0$, so $C=0$ and we are done, so we may assume $B\neq 0$ so $B>0$.
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Clever step: For any $t\in \mathbb{R}$,
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$$
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0\leq \sum (a_j-t b_j)^2=\sum (a_j^2-2ta_jb_j+t^2b_i^2)=A-2tC+t^2B
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$$
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let $t=C/B$ to get $0\leq A-2(C/B)C+(C/B)^2B=A-\frac{C^2}{B}$
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to generalize this to $\mathbb{C}$, $A=\sum |a_j|^2,B=\sum |b_j|^2,C=\sum |a_j \bar{b_j}|$.
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</details>
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### Euclidean spaces
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Nothing much to say. lol.
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Normal dot product as inner product.
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read text... Theorem 1.37 |