Math 307 - Winter 2023

Friday, March 10, 2023

In the first part of Problem 2 on Assignment 3 you are asked to create a function that would produce iterations of the quadratic type 2 curve. In the first picture below I show the 0th iteration in blue and the 1st iteration in red. I emphasize the points that are used. You do not need to do this on your plots. In the second picture below I show the 1st iteration in blue and the 2nd iteration in red. The large picture is the fourth iteration of the quadratic type 2 curve.

In the second part of Problem 2 you are ask you to create a function that would produce iterations of the Cesaro fractal which depends on angle $\alpha$. The four pictures below show the 0th, 1st, 2nd and the 3rd iteration of the Cesaro fractal with the small angle $\alpha = \pi/16$.

Below is an animated gif file that cycles through the 1st, 2nd, 3rd, 4th, 5th and the 6th iteration of the Cesaro fractal and, within each of the iterations cycles through all the angles starting from $\pi$, proceeding towards $0$ and then back to $\pi$ in steps of $\pi/50$. The animation starts by the first iteration and cycles through angles from $\pi$ to $0$ and back to $\pi$. This is repeated for 2nd, 3rd, 4th 5th and 6th iteration. For each iteration there are 101 pictures.

Place the cursor over the image to start the animation.

Thursday, March 2, 2023

As an introduction to the construction of the cycloid and the cardioid we studied the unit circle and its tangents today. A convenient tool in this construction is Mathematica's concept of pure function.

Pure Function. In the preceding exercises we often named them for further use. However, this is not always practical. Mathematica allows us to use the functions as if they were named without naming them. For example the square function is not named in Mathematica, but we can write (#^2)&[7] to get the square of 7. In some sense, we use (#^2)& as a name of the square function. Instead of naming the argument we are giving it a "place holder" # and we finish the definition with &.

The full power of pure function is realized in combination with the command Map[]. For example Map[(#^2)&,{a,b,c}] gives {a^2,b^2,c^2}. Thus, Map[] applies the specified function to the given list. Alternative shortcut form for Map[] is /@. In the previous simple example it would be (#^2)&/@{a,b,c}. This is extremely useful when we have a complicated function and a large list with hundreds of entries.
Next we play with the unit circle and its tangent lines. In the spirit of the above introduced concept of pure function, I define a tangent line to the unit circle as a pure function:
```
Line[{
     {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]},
         {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}
          }]&
```
To understand this definition you need to think of points as vectors. {Cos[#], Sin[#]} is the position vector of a point on the unit circle. The vector {-Sin[#], Cos[#]} is orthogonal to the specific position vector. In fact, this definition defines not a tangent line, but a finite line segment of length 20 joining the points {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]} and {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]} which are outside of the picture below. The midpoint of this line segment is {Cos[#], Sin[#]}, the point where the line segment touches the circle.

Since we used the pure function to make the above image, it is easy to Map[] the function to a range of values to get a more interesting picture
```
Line[{
  {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]},
    {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}
        }]&/@Range[0,2 Pi,(2 Pi)/5]
```
Replacing (2 Pi)/5 with Pi/128 we get an interesting image

Or, changing the domain for the range from 0,2 Pi with 0,(3/2) Pi we get an interesting image

Or, changing the domain for the range to 0,Pi we get an interesting image
In this item I present some variations on the preceding item. Extending tangents from the touching point in one direction we get Or, shortening the length of the touching line segments And some artistic images inspired by this theme
If you think that I was just playing in the preceding items, just as an option to consider, I did not. The preceding considerations naturally extend towards the construction of the involute of the unit circle.

Think of the circle below as an idealized roll of scotch tape. How is tape being unwrapped in a geometrically ideal way?

To answer the above question, we use the idea from the preceding work. Say we unwrap a piece of the tape of length $\pi/3.$ This is the corresponding geometrically ideal picture (click on the image to start an animation of the unwrapping of one full layer of tape)

The involute of the unit circle is the curve formed by the trace of the blue point in the preceding animation. In Mathematica we need to apply Line[] to the list of all points which are colored dark blue in the previous animation. For the animation hover cursor over the image.
Mathematica's command which created the complete involute of the unit circle in the preceding image is
```
{RGBColor[0, 0, 0.7], Thickness[0.003],
 Line[({Cos[#], Sin[#]} - # {-Sin[#], Cos[#]})&/@Range[0,3.28 Pi,0.02]]}
```
The corresponding traditional vector formula for the circle involute is: \[ \bigl( \cos(t) + t \sin(t), \sin(t) - t \cos(t) \bigr) \quad \text{with} \quad t \geq 0. \]
It is quite fascinating that the involute of the unit circle is related to the family of circles which are presented below as teal circles in the next image:
Mathematica's command for the teal circles is
```
{RGBColor[0, 0.6, 0.6], Thickness[0.001],
         Circle[{Cos[#], Sin[#]},#]& /@ Range[0,(5-1/6) Pi,Pi/48]}
```
In the next animation I present only the circles. You can see how the circles on their own recreate the involute. In fact, the involute is the envelope of this family of circles. This means that the involute is tangent to each of the circles in this family.

In the last scene of the preceding animation there are 233 circles. Below I show the image of 77 circles and even with fewer circles one can see the envelope phenomena of the involute.

Below I show the image of 38 circles and even with fewer circles one can see the envelope phenomena of the involute.
Another variation on the involute and the family of circles whose envelope is the involute is to present the circles and the involute as a 3d-image. We lift each circle at the level $z$ which equals the radius of the circle. In this way we form a surface in the image below:
```
Show[ParametricPlot3D[{Cos[z] + z Cos[t], Sin[z] + z Sin[t], z},
  {t,0,2 Pi}, {z,0,4 Pi}, Mesh -> {0, 4*24}, MeshStyle -> {None, RGBColor[0,0.6,0.6]},
  PlotPoints -> {100, 100}, PlotStyle -> {Opacity[0.2]}],
 ParametricPlot3D[{Cos[z] + z Sin[z], Sin[z] - z Cos[z], z},
 {z,0,4 Pi}, PlotPoints -> {200}, PlotStyle -> {RGBColor[0, 0, 0.7]}],
 BoxRatios -> {1, 1, 1}, Boxed -> False, Axes -> False, ImageSize -> 800]
```
The following animation illustrates the process of creating the surface described above:

To verify that the calculations for the 3d-image were correct, we view the 3d-object from bird's eye view

Wednesday, March 1, 2023

In Problem 2 on Assignment 2 you are asked to explore two famous curves cycloid and cardioid. First we discuss a cycloid.
Wikipedia's definition of a Cycoid is: A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slipping. One could demonstrate a cycloid in a dark room by attaching a small light to a rolling wheel and taking a long exposure photography of the motion. This is what American photographer Berenice Abbott did to create her photograph Cycloid:

You can read more about Berenice Abbott and her art at MoCP and Mount Holyoke College
In this item I will illustrate the derivation of the parametric equations for the cycloid which you need to implement in the animations in Mathematica. In the following derivation the vectors $\mathbf{i}$ and $\mathbf{j}$ are the unit coordinate vectors $\displaystyle \left[\begin{array}{c} 1 \\ 0 \end{array} \right]$ and $\displaystyle \left[\begin{array}{c} 0 \\ 1 \end{array} \right]$, respectively. In Mathematica notation the vector $\mathbf{i}$ is the pair {1,0} and the vector $\mathbf{j}$ is the pair {0,1}.
Let $t\in [0,\pi].$ The position vector of the gray center of the circle in the above pictures is \[ t \, \mathbf{i} + \mathbf{j}. \] If the unit circle was positioned at the origin, then the position vector of the red point on the unit circle would be \[ \bigl(\cos(-t -\pi/2) \bigr) \, \mathbf{i} + \bigl(\sin(-t -\pi/2) \bigr) \mathbf{j} = - (\sin t ) \, \mathbf{i} - (\cos t ) \mathbf{j}. \] Consequently, the position vector of the red point on the unit circle in the above pictures is \[ \bigl( t - \sin t \bigr) \, \mathbf{i} + \bigl(1 - \cos t \bigr) \mathbf{j}. \] In Mathematica notation the vector $\mathbf{i}$ is the pair {1,0} and the vector $\mathbf{j}$ is the pair {0,1}. Thus, the parametric equations of the cycloid in Mathematica notation is (t - Sin[t]) {1,0} + (1 - Cos[t]) {0,1} = {t - Sin[t] , 1 - Cos[t] } The following animation might help you to understand the above derivation.
Place the cursor over the image to start the animation.
In the first part of Problem 2 I ask you to reproduce the following animations.
Place the cursor over the image to start the animation.

Place the cursor over the image to start the animation.

Place the cursor over the image to start the animation.

In the second part of Problem 2 I ask you to explore a Cardioid.

Here is a derivation of the equation of a cardioid.

Click through several helpful images.

The origin $O = (0,0)$ of the coordinate system is the gray point. Denote by $C$ the black point in the picture.
Both circles in the picture are unit circles, the gray one centered at $O,$ the black one centered at $C.$
The angle at the gray point marked by the blue arc between the green and the blue point measures $t$ radians. That is, the length of this blue arc is $t.$
There are two other angles in the picture which also measure $t$ radians: the angle marked by the blue arc between the blue and the red point and the angle marked by the black arc between the blue point and the purple point.
The green point is $(1,0),$ the blue point is $(\cos t, \sin t),$ the black point $C$ is $(2\cos t, 2\sin t),$ the purple point is $(2\cos t-1, 2\sin t).$
The goal is to determine the coordinates of the red point.
Denote by $\mathbf{a}$ the unit vector whose tail is the black point and whose head is the red point. Based on the previous items we have \begin{align*} \mathbf{a} & = \bigl(\cos( \pi+2 t ) \bigr) \mathbf{i} + \bigl(\sin( \pi+2 t ) \bigr) \mathbf{j} \\ & = -\bigl(\cos(2 t ) \bigr) \mathbf{i} - \bigl(\sin(2 t ) \bigr) \mathbf{j}. \end{align*}
Denote by $\mathbf{c}$ the position vector of the black point. Then the position vector of the red point is \[ \mathbf{c} + \mathbf{a} = \bigl( 2 \cos t - \cos(2 t ) \bigr) \mathbf{i} + \bigl(2 \sin t - \sin(2 t ) \bigr) \mathbf{j}. \]
Thus the parametric equations for this cardioid are \begin{align*} x & = 2 \cos t - \cos(2 t ) \\ y & = 2 \sin t - \sin(2 t ) \end{align*}

In the Cardioid part of Problem 2 on Assignment 2 you need to reproduce two pictures from Wikipidia's Cardioid page I made an animation that unifies these two pictures. You should be able to produce something like this. If not, you can present one animation and one picture.
Place the cursor over the image to start the animation.
Once you have figured out how to construct a cardioid, you should be able to construct Generalized Cardioids. In this part of Problem 2 you need to produce generalized cardioids as shown in the animations below. These animated gif files are large. On a slow internet connection it takes a while for them to load.
- Below is the cardioid generated by a wheel with radius 1/2 rolling on a circle with radius 1.
  Place the cursor over the image to start the animation.
- Below is the cardioid generated by a wheel with radius 1/3 rolling on a circle with radius 1.
  Place the cursor over the image to start the animation.
- Below is the cardioid generated by a wheel with radius 2 rolling on a circle with radius 1.
  Place the cursor over the image to start the animation.
- Below is the cardioid generated by a wheel with radius 3/2 rolling on a circle with radius 1.
  Place the cursor over the image to start the animation.

Sunday, February 26, 2023

For better or worse, I am fascinated by the concept of "funny" trigonometric functions. I think it is for the better:
```
Fascination is a gift.
Let yourself be fascinated. 
```
So, I often look for new "funny" trigonometric functions. One idea that I explored was to combine the regular sine function and the trapezoidal sine function from Problem 1 on Assignment 1. What I mean by this is shown in the picture below. The yellow highlighted parts of the graph are from the trapezoidal sine function, and the cyan highlighted parts are from the standard sine function.

To plot the above function I used the Mathematica command Piecewise[], as follows


 Piecewise[{
   {Mod[x, 8 + 2 Pi],
    And[Mod[x, 8 + 2 Pi] >= 0, Mod[x, 8 + 2 Pi] < 1]},
   {1 + Sin[Mod[x, 8 + 2 Pi] - 1],
    And[Mod[x, 8 + 2 Pi] >= 1, Mod[x, 8 + 2 Pi] < 1 + Pi/2]},
   {2, And[Mod[x, 8 + 2 Pi] >= 1 + Pi/2, Mod[x, 8 + 2 Pi] < 3 + Pi/2]},
   {1 + Sin[Mod[x, 8 + 2 Pi] - 3],
    And[Mod[x, 8 + 2 Pi] >= 3 + Pi/2, Mod[x, 8 + 2 Pi] < 3 + Pi]},
   {(4 + Pi) - Mod[x, 8 + 2 Pi],
    And[Mod[x, 8 + 2 Pi] >= 3 + Pi, Mod[x, 8 + 2 Pi] < 5 + Pi]},
   {-1 + Sin[Mod[x, 8 + 2 Pi] - 5],
    And[Mod[x, 8 + 2 Pi] >= 5 + Pi, Mod[x, 8 + 2 Pi] < 5 + (3 Pi)/2]},
   {-2, And[Mod[x, 8 + 2 Pi] >= 5 + (3 Pi)/2,
     Mod[x, 8 + 2 Pi] < 7 + (3 Pi)/2]},
   {-1 + Sin[Mod[x, 8 + 2 Pi] - 7],
    And[Mod[x, 8 + 2 Pi] >= 7 + (3 Pi)/2,
     Mod[x, 8 + 2 Pi] < 7 + 2 Pi]},
   {Mod[x, 8 + 2 Pi] - (8 + 2 Pi),
    And[Mod[x, 8 + 2 Pi] >= 7 + 2 Pi, Mod[x, 8 + 2 Pi] < 8 + 2 Pi]}
   }]

What I soon discovered is that Piecewise[] when used in plotting complicated objects is very slow, almost unusable. There are two solutions for this problem

Find a way to define the desired function without using Piecewise[], that is by using some other already defined Mathematica functions. This is a preferred option.
Create a compiled version of the desired function by using the Mathematica command Compile[].

Since I was not able to find a simpler form for the desired function, I used Compile[] as follows


 Clear[LoSin, LoCos]; LoSin = Compile[{{x, _Real}}, Piecewise[
   {
    {Mod[x, 4 + Pi], And[Mod[x, 4 + Pi] >= 0, Mod[x, 4 + Pi] < 1/2]},
    {1/2 + 1/2 Sin[2 (Mod[x, 4 + Pi] - 1/2)],
     And[Mod[x, 4 + Pi] >= 1/2, Mod[x, 4 + Pi] < 1/2 + Pi/4]},
    {1, And[Mod[x, 4 + Pi] >= 1/2 + Pi/4,
      Mod[x, 4 + Pi] < 3/2 + Pi/4]},
    {1/2 + 1/2 Sin[2 (Mod[x, 4 + Pi] - 3/2)],
     And[Mod[x, 4 + Pi] >= 3/2 + Pi/4, Mod[x, 4 + Pi] < 3/2 + Pi/2]},
    {(4 + Pi)/2 - Mod[x, 4 + Pi],
     And[Mod[x, 4 + Pi] >= 3/2 + Pi/2, Mod[x, 4 + Pi] < 5/2 + Pi/2]},
    {-(1/2) + 1/2 Sin[2 (Mod[x, 4 + Pi] - 5/2)],
     And[Mod[x, 4 + Pi] >= 5/2 + Pi/2,
      Mod[x, 4 + Pi] < 5/2 + (3 Pi)/4]},
    {-1, And[Mod[x, 4 + Pi] >= 5/2 + (3 Pi)/4,
      Mod[x, 4 + Pi] < 7/2 + (3 Pi)/4]},
    {-(1/2) + 1/2 Sin[2 (Mod[x, 4 + Pi] - 7/2)],
     And[Mod[x, 4 + Pi] >= 7/2 + (3 Pi)/4, Mod[x, 4 + Pi] < 7/2 + Pi]},
    {Mod[x, 4 + Pi] - (4 + Pi),
     And[Mod[x, 4 + Pi] >= 7/2 + Pi, Mod[x, 4 + Pi] < 4 + Pi]}
    }]];
LoCos = Compile[{{x, _Real}}, LoSin[x + 1 + Pi/4]];

The function that I showed at the beginning of the post is in fact not a "funny" sine function since its range is $[-2,2].$ Therefore, for the definitions of the funny trigonometric functions LoSin[] and LoCos[], I scaled both dependent and independent variables of the function given at the beginning. Below are the graphs of the newly defined "funny" trigonometric functions.
As for all "funny" trigonometric functions we explore the funny unit circle the funny sphere the funny torus when we deal with tori whose cross-section is not a circle it is interesting to rotate the cross-section as it makes its turn to form a torus and the funny vase

Tuesday, February 14, 2023

In the notebook TheBeautyOfTrigonometry_12.nb (You can find this notebook in our shared Dropbox directory \307_Files\2022\.) I explore trigonometric functions Cos[t] and Sin[t] and how they are used in different settings in Mathematics. The most important feature of the functions Cos[t] and Sin[t] is that they provide a parametrization of the unit circle. The animation below illustrates how this parametrization is obtained.
The main components of the animation are as follows
- The black circle is the unit circle.
- The values of t $\in [0,2\pi]$ are represented by orange lengths. In each scene there are three orange lengths: the horizontal length, the vertical length and the arc length on the unit circle.
- The values of Cos[t] are green horizontal line segments.
- The values of Sin[t] are blue vertical line segments.

Place the cursor over the image to start the animation.

The principle presented above is the general principle of parametric equations of curves in the plane. Given two functions (one green and one blue) of the same independent variable t, taken as the coordinates of a moving point, then, as the independent variable t changes, the point does trace a curve in the plane.

It is of interest to consider the converse question: given a curve in the plane, for example an octagon inscribed in the unit circle, find a green and a blue function that do trace that octagon. I will show how to find a variety of such functions below. This is a good source for "funny" trigonometric functions.
Inspired by the parametrization of the unit circle by the trigonometric functions Cos[t] and Sin[t], today in class I started a derivation of a parametrization of a regular pentagon inscribed in the unit circle. The same reasoning can be used to parameterize any regular $n$-gon, where $n$ is a positive integer greater than $2.$ The handwritten note below shows this derivation.

The derivation from the preceding item, implemented in Mathematica is as follows:


Clear[n, t, NgonCos, NgonSin];
NgonCos[t_, n_] :=
 Cos[\[Pi]/n] Sec[\[Pi]/n - Mod[t, (2 \[Pi])/n]] Cos[t];
NgonSin[t_, n_] :=
 Cos[\[Pi]/n] Sec[\[Pi]/n - Mod[t, (2 \[Pi])/n]] Sin[t]

The final product of this work are the graphs of the first nine regular $n$-gons inscribed in the unit circel: the equilateral triangle, the square, and the pentagon, the hexagon, the heptagon, the octagon, the nonagon, the decagon, the hendecagon, the dodecagon, see the Wikipedia page.

It is always good to explore several approaches to mathematical problems. Below I provide another way to parameterize a regular $n$-gon. Here $n$ is a positive integer greater than $2.$ The handwritten note below shows this derivation.

The formulas derived in the preceding item, implemented in Mathematica are as follows:


Clear[nNgCos, nNgSin, tt, nn];
nNgCos[tt_, nn_] := (1 - Mod[tt, 1]) Cos[(2 \[Pi] Floor[tt])/nn] +
  Mod[tt, 1] Cos[(2 \[Pi] (1 + Floor[tt]))/nn] ;
nNgSin[tt_, nn_] := (1 - Mod[tt, 1]) Sin[(2 \[Pi] Floor[tt])/nn] +
  Mod[tt, 1] Sin[(2 \[Pi] (1 + Floor[tt]))/nn]

The final product of this work are the graphs of the first nine regular $n$-gons inscribed in the unit circel: the equilateral triangle, the square, and the pentagon, the hexagon, the heptagon, the octagon, the nonagon, the decagon, the hendecagon, the dodecagon, see the Wikipedia page.

Monday, February 6, 2023

The Mathematica part of the class will start on Tuesday, February 7, 2023 in BH 215 at 2:00 pm.
The information sheet
We will use
To get started with Mathematica see my Mathematica page. Please watch the videos that are on my Mathematica page before the first class. Watching the movies is essential for being able to organize your homework notebooks well! I will not discuss the basics of notebook structuring in class.

Winter 2023
MATH 307: Mathematical computing
with Mathematica

Branko Ćurgus

Winter 2023 MATH 307: Mathematical computing with Mathematica

Branko Ćurgus

Winter 2023
MATH 307: Mathematical computing
with Mathematica