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pages/Math4111/Math4111_L10.md
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@@ -27,7 +27,7 @@ $K$ is compact if $\forall$ open cover, $\exists \{G_{\alpha_i}\}_{i=1}^n$ that
$\mathbb{R}$ is not compact since we can build a open cover $\{(x,x+2):x\in \mathbb{Z}\}$ such that we cannot find a finite subcover of $\mathbb{R}$ $\mathbb{R}$ is not compact since we can build a open cover $\{(x,x+2):x\in \mathbb{Z}\}$ such that we cannot find a finite subcover of $\mathbb{R}$
$\{1,2\}$ is compact let $ $\{G_{\alpha}\}_{\alpha\in A}$ be an open cover of $\{1,2\}$ $\{1,2\}$ is compact let $\{G_{\alpha}\}_{\alpha\in A}$ be an open cover of $\{1,2\}$
Ironically, $[0,1]$ is compact. This will follow from Theorem 2.40. Ironically, $[0,1]$ is compact. This will follow from Theorem 2.40.

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# Lecture 2
## Review?
$$
z_1=r_1(\cos\theta_1+i\sin\theta_1)=r_1\text{cis}(\theta_1)
$$
$$
z_2=r_2(\cos\theta_2+i\sin\theta_2)=r_2\text{cis}(\theta_2)
$$
$$
z_1z_2=r_1r_2\text{cis}(\theta_1+\theta_2)
$$
$$
\forall n\in \mathbb{Z}, z^n=r^n\text{cis}(n\theta)
$$
### De Moivre's Formula
$$
\forall n\in \mathbb{Z}, z^n=r^n\text{cis}(n\theta)
$$
## New Fancy stuff
Claim:
$$
\forall n\in \mathbb{Z}, z^{\frac{1}{n}}=\sqrt[n]{r}\text{cis}\left(\frac{1}{n}\theta\right)
$$
Proof:
Take an $n$th power, De Moivre's formula holds $\forall$ rational $k\in \mathbb{Q}$.
Example:
we calculate $1^{\frac{1}{3}}$
$$
1=\text{cis}\left(2k\pi\right)
$$
$$
1^{\frac{1}{3}}=\text{cis}\left(\frac{2k\pi}{3}\right)
$$
When $k=0$, we get $1$
When $k=1$, we get $\text{cis}\left(\frac{2\pi}{3}\right)=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$
When $k=2$, we get $\text{cis}\left(\frac{4\pi}{3}\right)=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$
#### Strange example
Let $p(x)=a_3x^3+a_2x^2+a_1x+a_0$ be a polynomial with real coefficients.
Without loss of generality, Let $a_3=1$, $x=y-\beta$
We claim $\beta=\frac{a_2}{3}$
$$
\begin{aligned}
p(x)&=(y-\beta)^3+a_2(y-\beta)^2+a_1(y-\beta)+a_0\\
&=y^3+\left(a_2-3\beta\right)y^2+\left(a_1-3\beta^2-2a_2\beta\right)y+\left(a_0-3\beta^3-3a_1\beta-a_2\beta^2\right)\\
\end{aligned}
$$
It's sufficient to know how to solve real cubic equations.
$$
q(x)=x^3+ax+b
$$
Let $x=w+\frac{c}{w}$
Solve
$$
\begin{aligned}
(w+\frac{c}{w})^3+a(w+\frac{c}{w})+b=0\\
w^3+3w\frac{c}{w}+3\frac{c^2}{w^2}+aw+\frac{ac}{w}+b=0\\
\end{aligned}
$$
We choose $c$ such that $3c+a=0$, $c=-\frac{a}{3}$
$$
\begin{aligned}
w^3+3\frac{c^2}{w}+b=0\\
w^6+bw^3+c^2=0\\
\end{aligned}
$$
Notice that $w^6+bw^3+c^2=0$ is a quadratic equation in $w^3$.
$$
w^3=\frac{-b\pm\sqrt{b^2-4c^3}}{2}
$$
So $w$ is a cube root of $\frac{-b\pm\sqrt{b^2-4c^3}}{2}$
$x=w+\frac{c}{w}=w-\frac{a}{3w}$
Example:
$$
p(x)=x^3-3x+1=0
$$
$a=-3$, $b=1$, $c=-\frac{a}{3}=-\frac{-3}{3}=1$
$$
\begin{aligned}
w^3&=\frac{-b\pm\sqrt{b^2-4c^3}}{2}\\
&=\frac{-1\pm\sqrt{1-4}}{2}\\
&=\frac{-1\pm\sqrt{3}i}{2}\\
\end{aligned}
$$
To take cube root of $w$,
$$
w^3=\text{cis}\left(\frac{2\pi}{3}+2k\pi\right)
$$
So
Case 1:
$$
w=\text{cis}\left(\frac{2\pi}{9}+\frac{2k\pi}{3}\right)
$$
It is sufficient to check $k=0,1,2$ by nth root of unity.
When $k=0$, $w=\text{cis}\left(\frac{2\pi}{9}\right)$
When $k=1$, $w=\text{cis}\left(\frac{8\pi}{9}\right)$
When $k=2$, $w=\text{cis}\left(\frac{14\pi}{9}\right)$
Case 2:
$$
w=\text{cis}\left(\frac{4\pi}{9}+\frac{2k\pi}{3}\right)
$$
When $k=0$, $w=\text{cis}\left(\frac{4\pi}{9}\right)$
When $k=1$, $w=\text{cis}\left(\frac{10\pi}{9}\right)$
When $k=2$, $w=\text{cis}\left(\frac{16\pi}{9}\right)$
So the final roots are:
$$
w+\frac{c}{w}=w+\frac{1}{w}
$$
$$
\text{cis}(\theta)+\frac{1}{\text{cis}(\theta)}=\text{cis}(\theta)+\text{cis}(-\theta)=2\cos(\theta)
$$
So the final roots are:
$$
2\cos\left(\frac{2\pi}{9}\right), 2\cos\left(\frac{8\pi}{9}\right), 2\cos\left(\frac{14\pi}{9}\right), 2\cos\left(\frac{4\pi}{9}\right), 2\cos\left(\frac{10\pi}{9}\right), 2\cos\left(\frac{16\pi}{9}\right)
$$
Remember $\cos(2\pi-\theta)=\cos(\theta)$
So the final roots are:
$$
2\cos\left(\frac{2\pi}{9}\right), 2\cos\left(\frac{8\pi}{9}\right), 2\cos\left(\frac{14\pi}{9}\right)
$$
### Compact
A set $K\in \mathbb{R}^n$ is compact if and only if it is closed and bounded. [Compact Theorem in Math 4111](https://notenextra.trance-0.com/Math4111/Math4111_L12#theorem-241)
If $\{x_n\}\in K$, then there must be some point $w$ such that every disk $D(w,\epsilon)$ contains infinitely many points of $K$. [Infinite Point Theorem about Compact Set in Math 4111](https://notenextra.trance-0.com/Math4111/Math4111_L11#theorem-237)
Unfortunately, $\mathbb{C}$ is not compact.
### Riemann Sphere and Complex Projective Space
Let $\mathbb{C}\sim \mathbb{R}^2\subset \mathbb{R}^3$
We put a unit sphere on the origin, and project the point on sphere to $\mathbb{R}^2$ by drawing a line through the north pole and the point on the sphere.
So all the point on the north pole is mapped to outside of the unit circle in $\mathbb{R}^2$.
all the point on the south pole is mapped to inside of the unit circle in $\mathbb{R}^2$.
The line through $(0,0,1)$ and $(\xi,\eta,\zeta)$ intersects the unit sphere at $(x,y,0)$
Line $(tx,ty,1-t)$ intersects $\zeta^2$ at $t^2x^2+t^2y^2+(1-t)^2=1$
So $t=\frac{2}{1+x^2+y^2}$
$$
\zeta=x+iy\mapsto \frac{1}{1-|\zeta|}^2(2Re(\zeta),2Im(\zeta),|\zeta|^2-1)
$$
$$
(\xi,\eta,\zeta)\mapsto \frac{\xi+i\eta}{1-\zeta}
$$
This is a homeomorphism. $\mathbb{C}\setminus\{\infty\}\simeq S^2$
#### Derivative of a function
Suppose $\Omega$ is an open subset of $\mathbb{C}$.
A function $f:\Omega\to \mathbb{C}$'s derivative is defined as
$$
f'(\zeta_0)=\lim_{\zeta\to \zeta_0}\frac{f(\zeta)-f(\zeta_0)}{\zeta-\zeta_0}
$$
$f=u+iv$, $u,v:\Omega\to \mathbb{R}$
How are $f'$ and derivatives of $u$ and $v$ related?
1. Differentiation and complex linearity applies to $f$
Chain rule applies
$$
\frac{d}{d\zeta}(f(g(\zeta)))=f'(g(\zeta))g'(\zeta)
$$
Polynomials
$$
\frac{d}{d\zeta}\zeta^n=n\zeta^{n-1}
$$

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# Lecture 2: Binary Representation
## Bits review
- 1 byte = 8 bits
### Converting between binary and decimal
$162_{10} = 10100010_{2}$
$$
\begin{aligned}
162_{10} &= 1 \cdot 2^7 + 0 \cdot 2^6 + 1 \cdot 2^5 + 0 \cdot 2^4 + 0 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 0 \cdot 2^0 \\
&= 128 + 32 + 2 \\
&= 162
\end{aligned}
$$
## Example Data representations
| C Data Type | Size (bytes 32bit) | Size (bytes 64bit) | x86-64 |
| ----------- | ----------------- | ----------------- | ----- |
| `char` | 1 | 1 | 1 |
| `short` | 2 | 2 | 2 |
| `int` | 4 | 4 | 4 |
| `long` | 4 | 8 | 8 |
| `float` | 4 | 4 | 4 |
| `double` | 8 | 8 | 8 |
| `long double` | - | - | 10/16 |
| `pointer` | 4 | 8 | 8 |
Same size if declared as `unsigned`
### Encoding Integers (w bits)
#### Unsigned Integers
$$
B2U(X)= \sum_{i=0}^{w-1} x_i \cdot 2^i
$$
Example:
$$
\begin{aligned}
B2U(01101) &= 0 \cdot 2^4 + 1 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 \\
&= 0 + 8 + 4 + 0 + 1 \\
&= 13
\end{aligned}
$$
$$
\begin{aligned}
B2U(11101) &= 1 \cdot 2^4 + 1 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 \\
&= 16 + 8 + 4 + 0 + 1 \\
&= 29
\end{aligned}
$$
#### Two's Complement
$$
B2T(X)= -x_{w-1} \cdot 2^{w-1} + \sum_{i=0}^{w-2} x_i \cdot 2^i
$$
Example:
$$
\begin{aligned}
B2T(01101) &= 0 \cdot 2^4 + 1 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 \\
&= 0 + 8 + 4 + 0 + 1 \\
&= 13
\end{aligned}
$$
$$
\begin{aligned}
B2T(11101) &= -1 \cdot 2^4 + 1 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 \\
&= -16 + 8 + 4 + 0 + 1 \\
&= -3
\end{aligned}
$$
#### Sign Bit
- For 2's complement, most significant bit is the sign bit
- 0 for positive
- 1 for negative
#### Numeric Ranges
- Assume you have a integer type that is 4 bits long
- Unsigned: 0 to 15
- $0b1111=15$
- 2's Complement: -8 to 7
- $0b1000=-8$
- Unsigned Values:
- $UMin=0=B2U(000\ldots 0)$
- $UMax=2^w-1=B2U(111\ldots 1)$
- 2's Complement Values
- $TMin=-2^{w-1}=B2T(100\ldots 0)$
- $TMax=2^{w-1}-1=B2T(011\ldots 1)$
- Other interesting values $-1=B2T(111\ldots 1)$
#### Values for different word sizez
| | W=8 | W=16 | W=32 | W=64 |
| --- | --- | --- | --- | --- |
| TMin | -128 | -32768 | -2147483648 | -9223372036854775808 |
| TMax | 127 | 32767 | 2147483647 | 9223372036854775807 |
| UMin | 0 | 0 | 0 | 0 |
| UMax | 255 | 65535 | 4294967295 | 18446744073709551615 |
## C Operators for bitwise operations
### Boolean algebra
And
|`&`| 0 | 1 |
| --- | --- | --- |
| 0 | 0 | 0 |
| 1 | 0 | 1 |
Example:
```c
int a = 0b1010;
int b = 0b1100;
int c = a & b; // should return 0b1000
```
Or
|`\|`| 0 | 1 |
| --- | --- | --- |
| 0 | 0 | 1 |
| 1 | 1 | 1 |
Example:
```c
int a = 0b1010;
int b = 0b1100;
int c = a | b; // should return 0b1110
```
Xor
|`^`| 0 | 1 |
| --- | --- | --- |
| 0 | 0 | 1 |
| 1 | 1 | 0 |
Example:
```c
int a = 0b1010;
int b = 0b1100;
int c = a ^ b; // should return 0b0110
```
Not
|`~`| 0 | 1 |
| --- | --- | --- |
| | 1 | 0 |
Example:
```c
int a = 0b1010;
int c = ~a; // should return 0b0101
```
#### Imagine as set operations
- `&` is intersection
- `|` is union
- `^` is exclusive or (symmetric difference)
- `~` is complement
#### Logic operators on C
- `&&` is and
- `||` is or
- `!` is not
Interesting properties:
- View `0` as `false` and any non-zero value as `true`
- Always returns `0` or `1`
- Early termination: if the first operand is `0`, the second operand is not evaluated
Example:
```c
int a = 0x69;
int b = 0x55;
int c = a && b; // should return 0x01
int d = a || b; // should return 0x01 (the program will not check b at all since a is non-zero by early termination)
int e = !a; // should return 0x00
int f = !!a; // should return 0x01
int g = !0; // should return 0x01
int *p = NULL;
bool should_access = p && *p; // (avoid null pointer access, returns 0 if p is NULL, otherwise returns true if *p is non-zero)
```
#### Using bit masks
```c
// goal: compute val mod x and x is a power of 2
unsigned int val = 137; // some value to take mod of
unsigned int x = 16; // x is a power of 2
unsigned int mask = x - 1; // mask is a bit mask that is all 1s except for the least significant bit
unsigned int mod = val & mask; // mod is the result of val mod x
```
#### Shift operations
- `<<` is left shift
- Shift bit-vector x left y positions
- `>>` is right shift
- Shift bit-vector x right y positions
- Logical shift: fill with 0s
- Arithmetic shift: fill with the sign bit
- Undefined behavior: shift by a number greater than or equal to the word size
Example:
| `x` | `0b01100010` |
| --- | --- |
| `x<<3` | `0b00010000` |
| Logical shift `x>>2` | `0b00011000` |
| Arithmetic shift `x>>2` | `0b00011000` |
For negative numbers:
| `x` | `0b10100010` |
| --- | --- |
| `x<<3` | `0b00010000` |
| Logical shift `x>>2` | `0b00101000` |
| Arithmetic shift `x>>2` | `0b11101000` |
#### Pop count function
- How do you implement a pop count function in a 4-byte memory?
Trivial way:
```c
# define MASK 0x1;
int pop_count(unsigned int x) {
// does not work for negative numbers
int count = 0;
while (x!=0) {
if (x & MASK) count++;
x >>= 1;
}
return count;
}
```
#### Casting Between Signed and Unsigned Integers in C
Constants
- By default, constants are signed
- To make a constant unsigned, add the `U` suffix
```c
unsigned int a = 0x1234U;
```
Casting
- Explicitly cast to a different type
```c
int tx,ty;
unsigned int ux,uy;
tx = (int) ux;
uy = (unsigned) ty;
```
- Implicit casting also occurs via assignments and procedure calls
```c
tx = ux;
pop_count(tx); // popcount is a built-in function that returns the number of 1s in the binary representation of x (unsigned int)
```
#### When should I use unsigned integers?
- Don't use just because the number are non-negative
- Easy to make mistakes
```c
unsigned i;
for (i = cnt-2; i < 0; i++) {
// do something
}
```
If `cnt=1` then `i` will be `-1` and the loop will not terminate in short time. LOL.
- Can be very subtle
```c
#define DELTA sizeof(int) // sizeof(int) returns unsigned
int x = 0;
for (int i = CNT; i-DELTA >=0; i-=DELTA) {
// do something
}
```
The expression `i-DELTA >= 0` will be evaluated as `unsigned` and will not terminate.
#### Code Security Example
You can access the kernel memory region holding non user-accessible data. if you give negative index to the array, it will access the kernel memory region by interpreting the negative index as an unsigned integer.
## Change int size
### Extension
- When operating with types of different widths, C automatically perform extension
- Converting from smaller to larger type is always safe
- Given w-bit integer `x`,
- Convert `x` to `w+k` bit integer with the same value
- Two different types of extension
- zero extension: use for unsigned (similar to logical shift)
- sign extension: use for signed (similar to arithmetic shift)
### Truncation
- Task:
- Given w-bit integer `x`,
- Convert `x` to `k` bit integer with the same value
- Rule:
- Drop high-order `w-k` bits
- Effect:
- can change the value of `x`
- unsigned: mathematical mode on `x`
- signed: reinterprets the bit (add -2^k to the value)
#### Code puzzle
what is the output of the following code?
```c
unsigned short y=0xFFFF;
int x = y;
printf("%x", x); /* print the value of x as a hexadecimal number */
```
The output is `0x0000FFFF` it will try to preserve the value of `y` by sign extending the value of `y` to `x`.