Math 307 - Winter 2025

Friday, March 14, 2025

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.

And an animated gif with a different esthetics. It cycles through the 1st, 2nd, 3rd, 4th, and the 5th iteration of the Cesaro fractal

Place the cursor over the image to start the animation.

I post some useful calculations related to the Cesaro fractal in Problem 2 on Assignment 3. The construction of the Cesaro fractal is similar to the construction of the Koch snowflake that we did in class.

But, since the Cesaro fractal contains a "sawtooth" which depends on the angle, which I named $\alpha$, to determine the locations of points which form the "sawtooth", I needed to use trigonometry. See the handwritten notes below.

The related notebooks that I created in class are
- 20250313_Koch_snowflake.nb,
- 20250314_Cesaro.nb .
located in our shared Dropbox directory \307_Files\2025\.

Below is the Mathematica code which follows the same logic that we used in the construction of the Koch snowflake. The main difference in the code below is that, in addition to two points as variables, the function below also depends on the angle $\alpha$, which, in the code, I call al.
```
Clear[CesaroM,al];
CesaroM[{pA_, pB_}, al_] := 
{pA, (1 - 1/(2 + 2 Sin[al/2])) pA + 1/(2 + 2 Sin[al/2]) pB,
    1/2 pA + 1/2 pB + Cos[al/2] /(2 + 2 Sin[al/2]) {{0, -1}, {1, 0}}.(pB - pA),
                           1/(2 + 2 Sin[al/2]) pA + (1 - 1/(2 + 2 Sin[al/2])) pB, pB}
```
Always test your commands with specific values.
```
CesaroM[{{-3, 0}, {3, 0}}, Pi/3]
```
And with Manipulate[] as appropriate.
```
Manipulate[Graphics[{
   {Line[CesaroM[{{-3, 0}, {3, 0}}, al]]}
   },
  PlotRange -> {{-4, 4}, {-1, 4}}
  ], {{al, Pi/3}, 0, Pi}]
```
The above Mathematica commands you can copy-paste directly into a Mathematica notebook.

Thursday, March 13, 2025

Today I was not careful saving the notebook that I wrote during the class in which I constructed the Koch Snowflake fractal. This construction is useful for Problem 2 on Assignment 3.

I lost the notebook because I did not save my work while presenting the construction. At the end of class, I asked Mathematica to perform a challenging calculation. I knew the calculation might take a long time, but I was hoping that I would be able to abort it if it ran into trouble. Unfortunately, Mathematica crashed, and I lost all my work. The lesson here is to save work often, especially before attempting something challenging.

After the class, I reconstructed this notebook. See
- 20250313_Snowflake_Koch.nb,
located in our shared Dropbox directory \307_Files\2025\.

Tuesday, March 11, 2025

Yesterday and today we discussed Problem 1 on Assignment 3, explorations around around Goldbach's conjecture. The summaries of the classroom presentations are in the files below.
- 20250311_Golbach_partitions.nb,
- 20250310_While_Goldbach.nb,
located in our shared Dropbox directory \307_Files\2025\ will provide enough ideas for you to create your own explorations.
In Problem 1 Part (b) we explore all distinct Goldbach partitions of even integers greater than $4$. In fact, we are interested only in the count of distinct Goldbach partitions for each even integer; call this count the Goldbach length of an even number. For example, \begin{align*} 32 &= 3+29 = 13+19, \\ 34 &= 3 + 31 = 5 + 29 = 11 + 23 = 17 + 17, \\ 36 & = 5+31 = 7+29 = 13+23 = 17 + 19. \end{align*} Here are the first eighteen Goldbach lengths displayed as a Mathematica list:
{{6, 1}, {8, 1}, {10, 2}, {12, 1}, {14, 2}, {16, 2},
{18, 2}, {20,2}, {22, 3}, {24, 3}, {26, 3}, {28, 2},
{30, 3}, {32, 2}, {34, 4}, {36, 4}, {38, 2}, {40, 3}}
Below is a plot of the first sixty values of the Goldbach lengths. Reproduce the image below. Below is a plot of the first ten thousand values of the Goldbach lengths. Reproduce the image below. Below is a plot of the first fifty thousand values of the Goldbach lengths. Reproduce the image below. Below is a plot of the first five hundred thousand values of the Goldbach lengths. I wanted to create my own version of the picture from Wikipedia. If you can, reproduce the image below.
In Problem 2 Part (b-4) we explore:
- the Goldbach lengths of the even numbers which are divisible by $6$ (colored red in the plots below)
- the Goldbach lengths of the even numbers which leave remainder $2$ when divided by $6$ (colored green in the plots below)
- the Goldbach lengths of the even numbers which leave remainder $4$ when divided by $6$ (colored blue in the plots below)
Below is a colored plot of the first sixty values of the Goldbach lengths. Reproduce the image below. Below is a colored plot of the first ten thousand values of the Goldbach lengths. Reproduce one of the two images below. Below is a colored plot of the first fifty thousand values of the Goldbach lengths. Reproduce one of the two images: one above, one below. Below is a colored plot of the first five hundred thousand values of the Goldbach lengths. I wanted to create a colored version of the picture from Wikipedia. If you can, reproduce the image below.
Clearly the pictures below indicate that the even numbers which are divisible by $6$ have more Goldbach partitions than those even numbers which are not divisible by $6.$ Also, the pictures indicate that there is no significant difference between the number of Goldbach partitions of the even numbers which leave remainder $2$ when divided by $6$ and the even numbers which leave remainder $4$ when divided by $6$. How to provide statistical evidence for these claims?
- One way is to use moving averages, see Wikipedia page for moving average. I wrote a new section about moving averages in the class file 20240607_A3P2_Goldbach_dust_moving_average.nb in our shared Dropbox directory Dropbox\307_Files\2024.
The picture below shows the moving averages (of thousand values) of the values of the Goldbach length for fifty thousand even integers.

Sunday, March 9, 2025

On Thursday and Friday, I presented some explorations of the function
```
Clear[ff, x]; 

ff[x_] := (Sin[x] + Tanh[x])/(1 + Sin[x] Tanh[x])
```
given in Problem 2 on Assignment 2. I had hoped to summarize these explorations in a post here, but other obligations filled my weekend. Instead, I hope that the content in the files
- 20250303_A2P2_Explore_functions.nb,
- 20250306_SinTanh.nb,
- 20250307_A2P2_more_on_SinTanh.nb, and
- SinTanh_explorations.nb,
located in our shared Dropbox directory \307_Files\2025\ will provide enough ideas for you to create your own explorations.

Saturday, March 1, 2025

On Friday, I created the notebook 20250228_Cardioid.nb in our shared Dropbox directory Dropbox\307_Files\2024.
Thinking about what we did related to generalized cardioids, I focused my attention to the families of circles associated with different generalized cardioids. In the notebook 20250228_Cardioid.nb that I created on Friday, we encoutered interesting families of circles. Earlier, I posted cardioids with the rolling circles with radiuses $1$, $1/2$ and $1/3$. Together with these cardioids I presented families of circles whose envelopes are the related cardioids. It occurred to me that it might be interesting to invent a transition between those families of circles. Below is an animation that I created showing transitions from the families of circles associated with the cardioids with the rolling circles with radiuses \[ 1, \ \ \frac{1}{2}, \ \ \frac{1}{3}, \ \ \frac{1}{4}, \ \ \frac{1}{5}, \ \ \frac{1}{6}, \ \ \frac{1}{7}, \ \ \frac{1}{8}, \ \ \frac{1}{9}, \ \ \frac{1}{10}, \ \ \frac{1}{11}, \ \ \frac{1}{12}. \]

If the radius of the rolling circle is \[ r = \frac{1}{k} \quad \text{with} \quad k \in \bigl\{1,2,3,4,5,6,7,8,9,10,11,12 \bigr\} \] the corresponding family of circles that you see in the pictures above is given by \[ \text{center:} \ \bigl( \cos(t), \sin(t) \bigr), \quad \text{radius:} \ 2 r \left| \sin\left(\frac{t}{2 r}\right) \right|, \qquad \text{with} \quad t \in [0,2 \pi]. \] So, such a family is conveniently created in Mathematica using the following pure function
```
Circle[{ Cos[#], Sin[#] }, 2 r Abs[ Sin[ #/(2 r) ] ]  ]&/@Range[0, 2 Pi, Pi/64]
```
A complete Manipulation[] in which you can experiment with different radiuses and different number of circles in the family is as follows
```
Manipulate[Graphics[{
   {
    {AbsoluteThickness[2], AbsoluteThickness[1],
     RGBColor[0.95 {1, 1, 1}], Circle[{0, 0}, 1]}
    },
   {
    RGBColor[0.75 {1, 1, 1}], Thickness[0.002],
    Circle[{Cos[#], Sin[#]}, 2 r Abs[Sin[#/(2 r)]]] & /@
     Range[0, 2 Pi, Pi/n]
    },
   {
    {RGBColor[.5, 0, 0], AbsoluteThickness[2],
     Line[{(1 + r) Cos[#] - r Cos[(1 + 1/r) #], (1 + r) Sin[#] -
          r Sin[(1 + 1/r) #]} & /@ Range[0, 2 Pi, Pi/128]]},
    {RGBColor[.5, 0, 0], AbsoluteThickness[2],
     Line[{(1 - r) Cos[#] + r Cos[(1 - 1/r) #], (1 - r) Sin[#] +
          r Sin[(1 - 1/r) #]} & /@ Range[0, 2 Pi, Pi/128]]}
    }
   },
  AspectRatio -> Automatic, PlotRange -> {{-3.1, 1.7}, {-2.7, 2.7}},
  Frame -> False, ImageSize -> 600],
 {{r, 1}, 1/Range[1, 12], ControlPlacement -> Top,
  ControlType -> Setter}, {{n, 32}, {4, 8, 16, 32, 48, 64, 96},
  ControlPlacement -> Top, ControlType -> Setter}]
```
The animation below is 4 minutes long. It starts from the classical cardioid, transitions through each of the twelve cardioids listed above, and then loops back through all of them back to the cardioid.

Place the cursor over the image to start the animation.

A complete Manipulation[] in which you can experiment with the above transition is as follows
```
Manipulate[Graphics[
  {
   (* enveloped circles  *)

   {RGBColor[0.85 {1, 1, 0}], Thickness[0.002],
    Circle[{Cos[#], Sin[#]},
       (1 - Mod[t, 1]) 2/Floor[t] Abs[Sin[(Floor[t] #)/2]] +
        Mod[t, 1]*2/Ceiling[t] Abs[Sin[(Ceiling[t] #)/2]]] & /@
     Range[0, 2 Pi, Pi/nn]}


   },
  PlotRange -> {{-3.1, 1.7}, {-2.7, 2.7}},
  Frame -> False, ImageSize -> 400, Background -> Black

  ], {{nn, 16}, {4, 8, 16, 32, 64}, ControlPlacement -> Top,
  ControlType -> Setter}, {t, 1, 12, ControlPlacement -> Right,
  ControlType -> VerticalSlider}]
```
The last animation positioned horizontally:

Place the cursor over the image to start the animation.
In this item I will demonstrate a smooth transition from the standard cardioid determined by the family of circles that it envelopes into the unit circle involute, also determined by the family of circles that it envelopes.

The animation below starts from the classical cardioid and transitions to the circle involute and loops back to the cardioid.

Place the cursor over the image to start the animation.

A complete Manipulation[] in which you can experiment with the above transition is as follows
```
Manipulate[Graphics[
  {(* enveloped circles  *)
   {RGBColor[0.65 {1, 1, 0}], Thickness[0.001], 
    Circle[{Cos[#], Sin[#]}, 
       c*# + (1 - c) Norm[{1, 0} - {Cos[#], Sin[#]}]] & /@ 
     Range[0, 2 (1 + c/1.4)
       Pi, Pi/nn]} 
   },
  PlotRange -> {{-4.8, 7.9}, {-6.4, 6.3}}, Background -> Black,
  Frame -> False, ImageSize -> 500
  ], {{nn, 16}, {4, 8, 16, 32, 48, 64}, ControlPlacement -> Top, 
  ControlType -> Setter}, {{c, 1}, 0, 1}]
```

Thursday, February 27, 2025

In one of the problems on the next Assignment 2, I will ask you to explore the Cardioid and its generalizations.

The following table presents a derivation of the equation of a cardioid. For optimal viewing, use a wide screen.

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 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 the animation below . 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 Assignment 2 you are asked to produce some of the 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.
For completeness, I post four more generalized cardioids. The animations below 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 inside of 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 inside of 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 "inside" of 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 "inside" of a circle with radius 1.
Place the cursor over the image to start the animation.

Wednesday, February 26, 2025

We previously discussed a highly useful concept in Mathematica: the pure function. In my code, pay attention to the occurrences of pure functions and how I use them to create many lines or many points or many circles.

Pure Function. In earlier exercises, we often assigned a name to a function for further use. However, this approach is not always practical. Mathematica allows us to use functions as if they were named without actually naming them. For example, the square function is not explicitly named in Mathematica, but we can calculate the square of 7 by writing (#^2)&[7]. In this case, we use (#^2)& as a reference to the square function. Instead of explicitly naming the argument, we use a placeholder #, and we conclude the definition with &.

Notice that, instead of explicitly naming the argument (independent variable), in a pure function we use # which is called a slot in Mathematica. It acts as a placeholder for the argument that will be passed to the function. The symbol & is called function operator in Mathematica. It indicates the end of a pure function definition. For example, in the pure function (#^2)& the expression (#^2) defines the function, and ampersand & marks the end of the function.

The full potential of a pure function is realized when combined with the Map[] command. For instance, Map[(#^2)&,{a,b,c}] yields {a^2,b^2,c^2}. Thus, Map[] applies the specified function to each member of the given list. An alternative shortcut form for Map[] is /@. Using this shortcut, the previous example becomes (#^2)&/@{a,b,c}. This approach is extremely beneficial when dealing with functions given with several expressions and large lists containing hundreds of entries.
On Assignment 2 I will ask you to explore two famous curves: the cycloid and the cardioid. First we discuss the 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
Cycloid. Derivation of the equation. 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} 
```
Or, simplified,
```
 
{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 one of the problems on Assignment 2 I will ask you to reproduce the following three 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.

Monday, February 24, 2025

What I present today is an introduction to the construction of the cycloid and the cardioid. Today we study the unit circle and its tangents.
Although, I did not introduce the concept of pure function today in class, I will use it here and I will introduce this concept tomorrow in class.
A convenient tool in Mathematica constructions is the 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. The following vector,

{Cos[#], Sin[#]}

is the position vector of a point on the unit circle. Instead of a variable name we put # . The vector

{-Sin[#], Cos[#]}

is orthogonal to the vector specified above. Always think of points as vectors. The following pair of points

{{Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]},  
             {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}}

are on the tangent line to the unit circle. This tangent line touches the circle at the point

{Cos[#], Sin[#]}

This is illustrated in the picture below:

The complete code for the above picture is

 
Graphics[
 {
  {RGBColor[0, 0.5, 0.5], Thickness[0.005], Circle[{0, 0}, 1]},
 {
  Thickness[0.0015], RGBColor[0.7, 0.7, 0.7], 
   Line[{
   {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]}, 
            {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}}] &[4 Pi/5]}
  }, 
  {
  PlotRange -> {{-5, 5}, {-5, 5}}, Frame -> True, 
  PlotRangeClipping -> True, Background -> Automatic, 
  ImageSize -> 800
  }
  ]

In the above code, notice how we applied pure function to produce a line:

 
   Line[{
   {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]}, 
            {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}}] &[4 Pi/5]

Since we used the pure function to create the above image, it is easy to Map[] the pure function to a range of values to get a more interesting picture. There are two ways to implement Map[]. One is

Map[
Line[{
  {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]},
    {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}}]&, Range[0,2 Pi,(2 Pi)/5]
]

The other way is using the abbreviation /@

Line[{
  {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]},
    {Cos[#], Sin[#]} - 10 {-Sin[#], Cos[#]}
        }]&/@Range[0,2 Pi,(2 Pi)/5]

The complete code for the picture below is. Notice how we made just a small change to the preceding complete code

 
Graphics[
 {
  {RGBColor[0, 0.5, 0.5], Thickness[0.005], Circle[{0, 0}, 1]},
  {Thickness[0.0015], RGBColor[0.7, 0.7, 0.7], 
   Line[
   {
   {Cos[#], Sin[#]} + 10 {-Sin[#], Cos[#]}, {Cos[#], Sin[#]} 
              - 10 {-Sin[#], Cos[#]}
              }]&/@ Range[(2 Pi)/5, 2 Pi, (2 Pi)/5]}
  },
 {PlotRange -> {{-5, 5}, {-5, 5}}, Frame -> True, 
  PlotRangeClipping -> True, Background -> Automatic, 
  ImageSize -> 800}]

Changing the step in the Range[] function, from (2 Pi)/5 to Pi/128, we get an interesting image below

Or, changing the interval in the Range[] function from the original interval 0,2 Pi to 0,(3/2) Pi we get an interesting image

Or, setting the interval in the Range[] function to be 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

Sunday, Februarry 23, 2025

On Friday in class I played with twisted triangular torus, called a toroid.
Below I present two animations which show two ways to construct the one-third-twisted triangular toroid.

In this item, the one-third-twisted triangular toroid is constructed by a rotating equilateral triangle. The triangle is rotated by the full circle, that is the angle $2\pi$.

Place the cursor over the image to start the animation.

The code that is used for the above picture (not the animation) is below:

Show[ParametricPlot3D[
  3 {Cos[s], Sin[s], 0} +
   NgonCos[t, 3] (Cos[s/3] {Cos[s], Sin[s], 0} + Sin[s/3] {0, 0, 1}) +
    NgonSin[t,3] (-Sin[s/3] {Cos[s], Sin[s], 0} + Cos[s/3] {0, 0, 1}), {t, 0,
   3}, {s, 0, 2 Pi}, PlotPoints -> {51, 201},
  PlotStyle -> {Opacity[0.9]}, Mesh -> False, Exclusions -> None,
  PlotRange -> {{-4, 4}, {-4, 4}, {-1.1, 1.1}}],
 ParametricPlot3D[
  3 {Cos[s], Sin[s], 0} +
   NgonCos[0, 3] (Cos[s/3] {Cos[s], Sin[s], 0} + Sin[s/3] {0, 0, 1}) +
    NgonSin[0,
     3] (-Sin[s/3] {Cos[s], Sin[s], 0} + Cos[s/3] {0, 0, 1}), {s, 0,
   2 Pi}, PlotPoints -> {201},
  PlotStyle -> {RGBColor[0, 0, 0.75], Thickness[0.006]},
  Exclusions -> None],
 ParametricPlot3D[
  3 {Cos[s], Sin[s], 0} +
   NgonCos[1, 3] (Cos[s/3] {Cos[s], Sin[s], 0} + Sin[s/3] {0, 0, 1}) +
    NgonSin[1, 3] (-Sin[s/3] {Cos[s], Sin[s], 0} + Cos[s/3] {0, 0, 1}), {s, 0,
   2 Pi}, PlotPoints -> {201},
  PlotStyle -> {RGBColor[0.95, 0, 0], Thickness[0.006]},
  Exclusions -> None],
 ParametricPlot3D[
  3 {Cos[s], Sin[s], 0} +
   NgonCos[2, 3] (Cos[s/3] {Cos[s], Sin[s], 0} + Sin[s/3] {0, 0, 1}) +
    NgonSin[2, 3] (-Sin[s/3] {Cos[s], Sin[s], 0} + Cos[s/3] {0, 0, 1}), {s, 0,
   2 Pi}, PlotPoints -> {201},
  PlotStyle -> {RGBColor[0, 0.85, 0], Thickness[0.006]},
  Exclusions -> None], ImagePadding -> {{0, 0}, {0, 0}},
 Boxed -> True, BoxStyle -> {Opacity[0]}, Axes -> False,
 ImageSize -> 1000, PlotRange -> {{-4, 4}, {-4, 4}, {-1.1, 1.1}}]

The above code uses the functions that we defined earlier:

Clear[NgonCos, NgonSin];
NgonCos[t_, nn_] := (1 - (t - Floor[t])) Cos[(2/nn) \[Pi] Floor[t]] + (t -
      Floor[t]) Cos[(2/nn) \[Pi] (1 + Floor[t])];
NgonSin[t_, nn_] := (1 - (t - Floor[t])) Sin[(2/nn) \[Pi] Floor[t]] + (t -
      Floor[t]) Sin[(2/nn) \[Pi] (1 + Floor[t])];

In this item, the one-third-twisted triangular toroid is constructed by a rotating line segment. The line segment is rotated by the three full circles, that is by the angle $6\pi$.

Place the cursor over the image to start the animation.

The code that is used for the above picture (not the animation) is below:

Show[ParametricPlot3D[
  3 {Cos[s], Sin[s], 0} +
   NgonCos[t, 3] (Cos[s/3] {Cos[s], Sin[s], 0} + Sin[s/3] {0, 0, 1}) +
    NgonSin[t, 3] (-Sin[s/3] {Cos[s], Sin[s], 0} + Cos[s/3] {0, 0, 1}), {t, 0,
   1}, {s, 0, 6 Pi}, PlotPoints -> {51, 201},
  PlotStyle -> {Opacity[0.9]}, Mesh -> False, Exclusions -> None,
  PlotRange -> {{-4, 4}, {-4, 4}, {-1.1, 1.1}}],
 ParametricPlot3D[
  3 {Cos[s], Sin[s], 0} +
   NgonCos[0, 3] (Cos[s/3] {Cos[s], Sin[s], 0} + Sin[s/3] {0, 0, 1}) +
    NgonSin[0, 3] (-Sin[s/3] {Cos[s], Sin[s], 0} + Cos[s/3] {0, 0, 1}), {s, 0,
   6 Pi}, PlotPoints -> {201},
  PlotStyle -> {RGBColor[0, 0, 0.75], Thickness[0.006]},
  Exclusions -> None], ImagePadding -> {{0, 0}, {0, 0}},
 Boxed -> True, BoxStyle -> {Opacity[0]}, Axes -> False,
 ImageSize -> 1000, PlotRange -> {{-4, 4}, {-4, 4}, {-1.1, 1.1}}]

The above code uses the functions that we defined earlier:

Clear[NgonCos, NgonSin];
NgonCos[t_, nn_] := (1 - (t - Floor[t])) Cos[(2/nn) \[Pi] Floor[t]] + (t -
      Floor[t]) Cos[(2/nn) \[Pi] (1 + Floor[t])];
NgonSin[t_, nn_] := (1 - (t - Floor[t])) Sin[(2/nn) \[Pi] Floor[t]] + (t -
      Floor[t]) Sin[(2/nn) \[Pi] (1 + Floor[t])];

Thursday, February 20, 2025

In the PNG picture below I explain the geometry that is behind the definition of NgonCos[] and NgonSin[] which I defined today in class. In the PNG picture below I used the function Mod[t,1]. It is just the function t - Floor[t].
Today, we introduced functions inspired by the trigonometric functions cosine and sine, but designed to draw regular ngons, for example equilateral triangle, square, pentagon, hexagon, heptagon, octagon, nonagon, decagon, hendecagon, dodecagon, and so on. The notebook that I created today you can find in our shared Dropbox directory \307_Files\2025\ under the name 20250220_More_plots.nb.
The most important feature of the functions Cos[t] and Sin[t] is that they provide a parametrization of the unit circle. An animation illustrating this feature was posted on Saturday.
The principle presented in the animation posted on Thursday 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.

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

Clear[NgonCos, NgonSin];
NgonCos[t_, nn_] := (1 - (t - Floor[t])) Cos[(2/nn) \[Pi] Floor[t]] + (t -
      Floor[t]) Cos[(2/nn) \[Pi] (1 + Floor[t])];
NgonSin[t_, nn_] := (1 - (t - Floor[t])) Sin[(2/nn) \[Pi] Floor[t]] + (t -
      Floor[t]) Sin[(2/nn) \[Pi] (1 + Floor[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 Regular Polygon webpage at Wolfram MathWorld.

Saturday, February 15, 2025

To appreciate the Beauty of Trigonometry, it is essential to understand the key feature of the trigonometric functions Cos[t] and Sin[t]:
- The trigonometric functions Cos[t] and Sin[t] provide a parametrization of the unit circle. But that is just the beginning—most importantly, a deep understanding of the sphere and torus is impossible without Cos[t] and Sin[t].
The animation below is my attempt to illustrate how the parametrization of the unit circle works.
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 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, if these functions define the coordinates of a moving point, then as the independent variable t varies, the point traces a curve in the plane.

Friday, February 14, 2025

Today in class, a student ask about the difference between the commands Graphics[ ] on one side and Plot[ ]; ParametricPlot[ ] on the other side. The answer that I gave in class is in the notebook 20250214_More_plots.nb in our shared Dropbox directory Dropbox\307_Files\2025.

While answering this question, I started playing with the graph of the sine function. I moved it around, placed it climbing along the $y$-axis, coloring each point differently. After the class, I continued playing with this idea and I made two animations that are shown below.

The code for the animations is in the same file. The code is not documented, so it might be cryptic. I might documented, time permitting. You can try asking LLM models whether they understand my code.

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.

Thursday, February 13, 2025 (updated)

It is important in life to orient oneself in the world of COLORS. To help with that, I wrote a webpage: Color Cube. And all illustrations on the Color Cube page are created in Mathematica.
Being able to navigate the world of C O L O R S is important. To help with this, I created a page: Color Cube. All illustrations on the Color Cube page were generated using Mathematica.
To explore graphics Mathematica's graphics functions, we studied the notebook that I wrote: TheBeautyOfTrigonometry.nb.
- You can download this notebook here.
- You can also find it in our shared Dropbox directory Dropbox\307_Files\2025.
- I also created a pdf printout of this notebook. It is a long pdf document; you can read it when you don't have access to Mathematica.
- Another way how you can use the pdf file from the preceding item is to take a snippet from a command that you do not understand, paste the snippet in DeepSeek and ask for an explanation. You can accompany the snippet with a prompt like this: "Can you please explain the Wolfram Mathematica command that I gave you in the attached snippet?"
- ChatGPT does not allow free accounts to upload images. But, when you are working on a Mathematica notebook, you can copy a Mathematica command, and paste it into ChatGPT as text and it will explain the command for you. Remember, in ChatGPT, to start a new line in a prompt, you press Shift+Enter.
- You can also try Copilot. I did not explore its strengths and weaknesses.
- Share your experiences in Discussions on Canvas.

Wednesday, February 12, 2025

On Tuesday, we emphasised the important role played by brackets in Mathematica. Pay attention to the parenthesis ( ), the square brackets [ ], the curly brackets, or braces: { }, the double square brackets [[ ]], and finally, the stared parenthasis (* *). To illustrate the double square brackets [[ ]], I talked about the Hall of Fame of Numbers. The Hall of Fame of Numbers is the set consisting from the five numbers: the number one, $1$, the number zero, $0$, the number pi, $\pi$, the number $e$, and the imaginary unit $i$. In Mathematica notation, I write this set as a list and name it HFM:
```
HFN = {1, 0, Pi, E, I}
```
With this notation, I illustrate the use of the double square brackets. To extract the third entry in the preceding list I write:
```
HFN[[3]] 
```
Next, I will write Mathematica code for a famous mathematical equation and invite you to recognize it:
```
HFN[[4]]^(HFN[[5]]*HFN[[3]]) + HFN[[1]] == HFN[[2]]
```
Which famous equality is hidden in Mathematica code above?
I was curious whether ChatGPT will recognize what is hidden in the Mathematica code above. My prompt and ChatGPT's response is below.

Saturday, February 8, 2025

The Mathematica part of the class will start on Monday, February 10, 2025 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. 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.
In the study of the computational software system Wolfram Mathematica, we will focus on symbolic and numerical calculations, visualization, and programming in Mathematica. But Mathematica is much more than that. To read more about Mathematica see An Elementary Introduction to the Wolfram Language.
As we begin the Mathematica portion of this class, a natural question arises: Why learn Mathematica? This is not an easy question to answer, but below are some key aspects that make Mathematica worth your attention.
- Mathematica is a powerful computational tool that can handle a wide range of mathematical problems, from basic algebra to advanced calculus and beyond.
- Unlike many other mathematical software systems, Mathematica excels in symbolic computation, allowing you to manipulate mathematical expressions and equations symbolically, closely resembling abstract mathematical thinking and structures.
- Mathematica offers powerful visualization tools, allowing you to create compelling graphics that enhance the understanding of mathematical concepts and effectively illustrate abstract mathematical results, fostering a varied and engaging mathematical learning experience.
- Mathematica is a high-level programming language that offers an extensive array of predefined functions, designed to simplify and streamline tasks that are commonly encountered in mathematical and technical computing. My impression is that this makes it more accessible and user-friendly compared to other mathematical software, enabling users to achieve complex objectives with greater ease.
- A lot of help is available on-line, for example Mathematica Stack Exchange, Wolfram Community and ChatGPT is quite good at offering help with Mathematica questions.
- Mathematica makes solving mathematical problems more efficient. How? For problems involving symbolic manipulation, we can perform symbolic manipulations in Mathematica exactly. For problems in which symbolic manipulations are inefficient, numerical approximations can be performed. Additionally, powerful graphical visualizations help guide the problem-solving process, offering deeper insights.
I have enjoyed the interaction between my mind and computers since I first started using them in the early 1990s. Recently, I came across a powerful metaphor for this relationship, expressed by Steve Jobs. In an old interview, he illustrated the usefulness of computers with a compelling analogy: "For me, a computer has always been a bicycle for the mind." That perfectly describes Mathematica and other modern tools like ChatGPT. Isn't that wonderful?
See the Bicycle for the Mind video. Or, a longer version.
I am convinced that your mathematical learning experience will be enriched if you engage in mathematical explorations using technology.
I recently encountered a fascinating geometric fact online: Imagine two squares where a vertex of the larger square is positioned at the center of the smaller square. Regardless of how these two squares are positioned, the area of their intersection remains constant. My description might not fully capture the beauty of this fact — indeed, an animation is worth a thousand words. How am I to make an animation? I must use technology. Here is an animation created in Mathematica.

Place the cursor over the image to start the animation.
I should have left it as an exercise for you to figure out why what I claim in the preceding item is true. But, the solution supported by a Mathematica animation might indicate why using technology can be powerful tool in communicating Mathematics.
Place the cursor over the image to start the animation.

Winter 2025
MATH 307: Mathematical computing
with Mathematica

Branko Ćurgus

Winter 2025 MATH 307: Mathematical computing with Mathematica

Branko Ćurgus

Winter 2025
MATH 307: Mathematical computing
with Mathematica