NoteNextra-origin/pages/Math4111/Math4111_L5.md

Lecture 5

Review

In each case, determine (with justification) whether the claim or its negation is the true statement.

(a) For all real numbers satisfying a<b, there exists an n\in \mathbb{N} such that a+1/n<b.

negation: \exists a<b, \forall n\in \mathbb{N} such that a+1/n\geq b.

By Archimedean property, the statement is true.

(b) There exists a real number x>0 such that x<1/n for all n\in \mathbb{N}.

The statement is ambiguous because we can arrange the statement in two ways.

\exists x\in \mathbb{R}_{>0} such that \forall n\in \mathbb{N},x\leq \frac{1}{n}

negation: \forall x\in \mathbb{R}_{>0}, \exists n\in \mathbb{N}, such that x\leq \frac{1}{n}.

The statement is true, let x=\frac{1}{n+1}.

New Materials

Continue on the theorem

Theorem 1.21

\forall x\in \mathbb{R}_{>0},\forall n\in \mathbb{N},\exist unique y\in \mathbb{R}_{>0} such that y^n=x.

(Because of this Theorem we can define x^{1/x}=y and \sqrt{x}=y)

Proof:

We cna assume n\geq 2 (For n=1,y=x)

Step 1 (uniqueness): If 0<y_1<y_2, then y_1^n<y_2^n (by properties of ordered field)

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.

1. To show E\neq \phi:

   If x\geq 1, then 1/2\in E.

   If x<1, then x\in E.

2. To show E is bounded above. We need to find an upper bound.

   If x\geq 1, then x\in E

   If x<1, then 1 \in E.

So we can let y=sup\ E

Step 2b (y^n\geq x) Suppose for contradiction y^n<x.

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.

(y+h)^n=y^n+ny^{n-1}h+{more\ terms}

We want ny^{n-1}h+{more\ terms}<x-y^n

Observe: If 0<a<b, then

\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}

The fact tells us (y+h)^n-y^n\leq n(y+h)^{n-1}h.

And we want n(y+h)^{n-1} h+y^n<x.

So want h<\frac{x-y^h}{n(y+h)^{n-1}}.

To achieve this, choosey any h>0 satisfying h<1 and h<\frac{x-y^h}{n(y+h)^{n-1}}

[For actual proof, see the text.]

Step 2c showing (y^n\leq x)

Suppose for contradiction y^n>x

Thoughts: Find k>0 such that (y-k)^n>x.

Then y-k is an upper bound for E, which contradicts the fact that y is the least upper bound of E.

y^n-(y-k)^n\leq ny^{n-1}k.

We want y^n-ny^{n-1}k\geq x.

So want k\leq \frac{y^n-x}{ny^{n-1}}

[For actual proof, see the text.]

Complex numbers

1. =\{a+bi:a,b\in \mathbb{R}\}.

Conjugate: z=a+bi,\bar{z}=a-bi.

Theorem 1.31 (see text)

Pure computational proof: boring...

z\bar{z}=a^2-(bi)^2=a^2+b^2

You can also use vector sum for representing operation in complex numbers.

Theorem 1.33 (see text)

More computation and still, boring...

some fun theorems:

• |Re\ z|\leq |z| (equal when no imaginary parts)
• |z+w|\leq |z|+|w| (equal when both z,w have no imaginary parts) (Triangle inequality)

Proof for |z+w|\leq |z|+|w|:

|z+w|^2=(z+w)(\overline{z+w})=(z+w)(\bar{z}+\bar{w})=|z|^2+|w|^2+z\bar{w}+\bar{z}w

Since

z\bar{w}+\bar{z}w\leq 2Re(z\bar{w})

(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|

Theorem 1.35 Cauchy-Schwarz inequality

If \vec{a},\vec{b}\in \mathbb{C}^n, then

|\vec{a}\vec{b}|^2\leq (\vec{a}\vec{a})(\vec{b}\vec{b})

Remark: The proof is very tricky.

To help us motivate the proof in text, let's consider the special case of real numbers.

(\sum a_j b_j)^2=(\sum a_j^2)(\sum b_j^2)

Proof for real numbers:

Let A=\sum a_j^2,B=\sum b_j^2, C=\sum a_j b_j, want to show C^2\leq AB

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.

Clever step: For any t\in \mathbb{R},

0\leq \sum (a_j-t b_j)^2=\sum (a_j^2-2ta_jb_j+t^2b_i^2)=A-2tC+t^2B

let t=C/B to get 0\leq A-2(C/B)C+(C/B)^2B=A-\frac{C^2}{B}

to generalize this to \mathbb{C}, A=\sum |a_j|^2,B=\sum |b_j|^2,C=\sum |a_j \bar{b_j}|.

Euclidean spaces

Nothing much to say. lol.

Normal dot product as inner product.

read text... Theorem 1.37