Dr Juan H Klopper

Let's kick things off by considering numbers and numbering systems and start off with the natural (counting) numbers, $ \mathbb{N} $. They start at $1$ and increment by $1$. That's great for adding, but what about subtracting?

To account for negative addition (subtraction) we introduce the integers, $ \mathbb{Z} $. They are $\dots -n,-n+1,-n+2,\dots -2,-1,0,1,2,\dots ,n-2,n-1,n,\dots $.

To account for division we require the rational numbers, $\mathbb{Q}$, which are the ratios of integers.

To complete the gaps left by the rational number we require numbers that are not ratios of integers, the socalled irrational numbers, $\mathbb{I}$ , such as $\pi$ and $\sqrt { 2 }$. Together, these numbers make up the real numbers, $\mathbb{R}$, presented geometrically on a real line $x$-axis on a Cartesian plane.

This does not encompass all numbers, though. We need to introduce complex numbers, $\mathbb{C}$. A complex number consist of a real part and an imaginary part. Both of these are represented as real numbers, but the imaginary part is multiplied with the imaginary number $i$, which is $\sqrt{-1}$.

*The algebraic structure in $\mathbb{R}^{2}$*. $\mathbb{R}^{2}$ represents ordered pairs of real numbers as in $\left(x,y\right)$. They can be visualized as points on a two-dimensional, Cartesian plane.

*zeta* $\left(\zeta\right)$, *xi* $\left(\xi\right)$, and *eta* $\left(\eta\right)$, such that $\zeta=\left(\xi , \eta \right) $.

In a simple example we have $$ z=\left( x,y \right) \\ \zeta =\left( \xi ,\eta \right) \\ z\pm \zeta =\left( x\pm \xi ,y\pm \eta \right) $$

We also define constant multiplication $$ cz = \left( cx,cy \right) $$

In [1]:

```
# Using direct importing and namespace abbreviation
from numpy import random
import matplotlib.pyplot as plt
```

In [2]:

```
# Plotting graphs directly in the notebook
%matplotlib inline
```

In [3]:

```
x_points = random.randint(-3, 3, 10) # Ten random integers ranging from -3 to 3
y_points = random.randint(-3, 3, 10)
```

In [4]:

```
# Creating a plot of size 10 by 10
plt.figure(figsize = (10, 10))
# Plotting ordered pairs of the x- and y-values above
# Using blue, round markers of size 12, with no lines connecting them
plt.plot(x_points, y_points, marker = "o", markerfacecolor = "blue", markersize = 12, linewidth = 0)
plt.hlines(0, xmin = x_points.min(), xmax = x_points.max(), label = "x-axis")
# Plotting the x- and y-axis
plt.vlines(0, ymin = y_points.min(), ymax = y_points.max(), label = "y-axis");
```

With $ \xi = 0 $ and $\eta = 0$ we have $$ \left| z \right| = \sqrt{{x}^{2}+{y}^{2}} $$

*open*, *connected*, and *interior points*.

Consider a fixed point in $\mathbb{R}^{2}$, call it ${z}_{0}$. Any (free to move) point $z$ near $ {z}_{0} $ creates a distance $r=\left| z-{ z }_{ 0 } \right| $ between them. If we fix $r$ we have a circle or **disk** around ${z}_{0}$.

For an **open** set of points in this disk, we do not include the rim $$ D\left( { z }_{ 0 };r \right) =\left\{ z\in \mathbb{ R }^{ 2 },\left| z-{ z }_{ 0 } \right|

**interior point** of $\Omega$ if we can create a disk around $z$ (by the method above, i.e. have a fixed $r>0$) if all the points that make up this disk falls entirely inside $\Omega$ $$ D\left( z;r \right) \subset \Omega $$

**open** set then excludes the boundary points. If we included the boundary (border) points in a subset (disk) we cannot create a (new) disk (with radius $r$) around such a border point such that all the points in this (new) disk fall within the original subset.

We also define any open set, $S$, minus one of its interior points, $S-\left\{z\right\}$, as open.

**disconnected** if there union is the empty set, $\emptyset$. This means that they have not point in common.

An open set is **connected** if and only if it is not disconnected. Another way to look at this is to say that an open set is connected if any two points in the set can be connected by a finite number of straight lines all going through points that are members of the set.

**bounded** subset in $\mathbb{R}^{2}$ is bounded if it can be contained within some disc with a finite radius, $r$. The upper half of the plane, for instance, cannot be a bounded set, as it moves off into infinity as $y\rightarrow \infty $. Similarly the graph of a function such as $y=\cos{x}$ or $\left| z \right|>r$ are not bounded. On the other hand we have examples of bounded sets such as the empty set, $\emptyset$, an open or closed set, a single point, any finite set of points, a subset of a bounded set and functions such as ${r}_{1}<\left|z-{z}_{0}\right|<{r}_{2}$.

**boundary point** by considering a point, $z$ on the edge of a disk, $S$. If we construct *any* disk around $z$ and some of the points in this new disk falls within S and some don't, then $z$ is a boundary point of $S$. Importantly, $z$ does not have to be an element of $S$. Think about $S$ as an open disk with elements only within a radius $

One half of the plane is an interesting case. It is an unbounded set, but with a boundary, i.e. the boundary of the unbounded set $y>0$ is $y=0$.

We can parametrize this curve. Introducing a new parameter (variable) such as $t$, we can create an interval $\left[0,2\pi \right]$, such that $, \le t \le 2 \pi$. A function in $t$ can then be written as $\alpha \left( t \right) = \left( x,y \right) = \left( r \cos{t}, r \sin{t} \right)$.

With $\left| \alpha \left({t} \right) \right| = r$ we have that every point $ \alpha \left( {t} \right) $ lies on the circle, $C=C\left( 0;r \right)$. A new notation for this would be $\alpha : \left[ 0, 2 \pi \right] \rightarrow C \left( 0;r \right)$.

**simple parametrization**. We want $ \alpha \left( {t}_{1} \right) = \alpha \left( {t}_{2} \right) $ to be the case only when $ {t}_{1} = {t}_{2}$. That means, we don't want the curves to cross over themselves. For a circle we can have a *crossing* at the start and end. This would consitute a closed loop.

A good example of a non-simple parametrization is one of our previous examples where we had a coefficient of $4$ for our parameter. For every $t \in \left[ a,b \right]$ we have $4$ values for $\alpha$.

In [5]:

```
import sympy as sym # Making using of namespace abbreviation
```

In [6]:

```
# Using Latex to print matehmatical notation
sym.init_printing(use_latex = "mathjax")
```

In [7]:

```
# Creating algebraic symbols
r, t = sym.symbols("r, t")
```

In [8]:

```
# Using the Derivative function which prints the derivative to the screen
# Arguments are function and variable with respect to which derivative is taken, i.e.
# (function, t)
sym.Derivative(r * sym.cos(2 * t), t)
```

Out[8]:

In [9]:

```
sym.Derivative(r * sym.sin(2 * t), t)
```

Out[9]:

In [10]:

```
# Calculating the actual derivative using .doit()
sym.Derivative(r * sym.cos(2 * t), t).doit()
```

Out[10]:

In [11]:

```
# Calculating the derivative using sym.diff(function,variable)
sym.diff(r * sym.sin(2 * t), t)
```

Out[11]:

In [12]:

```
# The integral
sym.Integral(sym.sqrt(((sym.diff(r * sym.cos(2 * t), t))**2) +
((sym.diff(r * sym.sin(2 * t), t))**2)), (t, 0, 3 * sym.pi / 2))
```

Out[12]:

In [13]:

```
# Actual calculation
sym.integrate(sym.sqrt(((sym.diff(r * sym.cos(2 * t), t))**2) +
((sym.diff(r * sym.sin(2 * t), t))**2)), (t, 0, 3 * sym.pi / 2))
```

Out[13]:

`.simplify()`

). The answer is thus $ 3 \pi {r}$.

Let's do this example $$ f \left( x \right) = {x}^{2}, x \in \left[ 0,1 \right] $$

In [14]:

```
sym.integrate((1/2) * ((sym.sec(t))**3), (t, 0, sym.atan(2))).simplify()
```

Out[14]:

In [15]:

```
from sympy.physics.vector import ReferenceFrame, dot
C = ReferenceFrame("C")
```

In [16]:

```
U = 3 * C.x + 7 * C.y
U
```

Out[16]:

In [17]:

```
V = -3 * C.x + 8 * C.y
V
```

Out[17]:

In [18]:

```
dot(U, V)
```

Out[18]:

Let's see how that looks symbolically.

In [19]:

```
u1, u2, v1, v2 = sym.symbols("u_1 u_2 v_1 v_2")
```

In [20]:

```
U = u1 * C.x + u2 * C.y
V = v1 * C.x + v2 * C.y
```

In [21]:

```
U
```

Out[21]:

In [22]:

```
V
```

Out[22]:

In [23]:

```
dot(U, V)
```

Out[23]:

It is also very easy using this notation to get the magnitude of $U$.

In [24]:

```
U.magnitude()
```

Out[24]:

So, we have a Jordan domain $ \Omega $, with a boundary, sometimes denoted as $ \partial \Omega $. We denote $ \Gamma $ as a Jordan curve that forms part of the boundary.

In [ ]:

```
```