Spring 2022
MATH 307: Mathematical computing
with Mathematica

Branko Ćurgus

Friday, June 3, 2022

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• Today we worked on Problem 2 on Assignment 3. The notes are in the notebook 20220602_A3P2.nb. As always, this notebook is in the folder 307_Files\2022 in Dropbox. We looked at the count of all Goldbach partitions of a specific even positive integer; call this count the Goldbach length of a positive even number. Here are the first Goldbach lengths:
  {{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}}
  This is not a very eventful list. Let us just verify that the Goldbach length of $34$ is indeed $4:$ \[ 34 = 3 + 31 = 5 + 29 = 11 + 23 = 17 + 17. \] However, when we calculate the corresponding list for the first five hundred thousand positive even integers one gets a beautiful picture which is called the Goldbach dust. This beautiful picture is at Wikipedia's page for Goldbach's conjecture. Here I made a picture of the first one hundred thousand positive even integers, but I beautified this picture with four trailing statistics about the Goldbach lengths. The dark blue is the trailing average for the previous 5000 Goldbach lengths. The purple is the trailing median for the previous 5000 Goldbach lengths. The dark green is the trailing minimum for the previous 5000 Goldbach lengths. The orange is the trailing maximum for the previous 5000 Goldbach lengths.

Below are Goldbach lengths of 500000 positive even integers. I just wanted to create my own version of the picture from Wikipedia.

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Tuesday, May 31, 2022

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• Today we discussed Problem 3 on Assignment 3, see 20220531_A3P3.nb. As an example of a fractal we constructed a Koch snowflake. As a more complicated example we looked at my notebook PythagorasTree.nb which is also in \307_Files\2022\. If you do not have access to our shared Dropbox file you can look at the pdf printout of notebook PythagorasTree.nb.

• Logan suggested that we do Sierpinski triangle as an exercise. Here is my construction of the first several iterations of the Sierpinski triangle. The function below, when combined with Nest[], on an equilateral triangle will result in the first iterations of the Sierpinski triangle.

  Clear[MoreTList, lsttri];
  MoreTList[lsttri_] :=
   Flatten[{{#[[1]], (1/2) (#[[1]] + #[[2]]),(1/2) (#[[1]] + #[[3]])},
                   {(1/2) (#[[1]] + #[[2]]), #[[2]], (1/2) (#[[2]] + #[[3]])},
                       {(1/2) (#[[1]] + #[[3]]), (1/2) (#[[2]] + #[[3]]), #[[3]]}} & /@ lsttri, 1]
  

  In the command below choose nn is the desired level of iterations of the Sierpinski triangle.

  nn=7; Graphics[{
    Polygon[#] & /@Nest[MoreTList[#] &,
      N[{{{Cos[7 Pi/6], Sin[7 Pi/6]},
               {Cos[-Pi/6], Sin[- Pi/6]}, {0,1}}}], nn]
    },
   Frame -> True, FrameStyle -> GrayLevel[0.8],
   FrameTicks -> {{{}, {}}, {{}, {}}},
   GridLines -> {Range[-1, 1, .01], Range[-2, 2, .01]},
   GridLinesStyle -> GrayLevel[0.9],
   PlotRange -> {{-0.9, 0.9}, {-0.55, 1.03}}, ImageSize -> 800]
  

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Monday, May 30, 2022

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• In the first part of Problem 3 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 3 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.

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Saturday, May 28, 2022

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• Today I will write up a command that I introduced on Friday working on Problem 3 on Assignment 2, see 20220526_A2P3.nb. On Friday we explored a setting in which we created a large list of trigonometric expressions some of which were rational numbers. Our goal was to recognize which expressions were rational numbers. In one case we produced a list of $9! = 362880$ trigonometric expressions and we wanted to select those of the expressions which simplify to a rational number. Since Mathematica was slow to FullSimplify[] this large number of expressions we needed to invent another way of deciding which expressions are rational, which are not.
• I will illustrate the method that we used on a simple small example. First I will produce a test list of $200$ random real numbers some of which will be rational, some of which will be irrational.

  test = If[# < 0,
                RandomInteger[{1, 200}]/RandomInteger[{1, 200}],
                     Sin[RandomInteger[{1, 1000}]]] & /@
             (RandomReal[{-1, 19}] & /@Range[200]);
  

  The above list test consists of $200$ real numbers, of which approximatelly $10$ are rational. As given, it is easy to select the rational numbers

  Select[test, Element[#, Rationals] &]
  

  However, let us convert test into its quite accurate numerical values by setting

   ntest = N[test, 40];
  

  Can we recover the original rational values from the list ntest?

  The command below does exactly that:

   Rationalize[#, 10^(-7)] & /@
   Select[ntest, (Abs[(Rationalize[#, 10^(-7)] - #)] < 10^(-20)) &]
  

  The idea in the Select[] criteria (Abs[(Rationalize[#, 10^(-7)] - #)] < 10^(-20)) & is that we ask Mathematica to Rationalize[] each of the decimal numbers in ntest with accuracy $10^{-7}.$ We expect that the rational numbers will be perfectly rationalized and that the irrational numbers will be rationalized with an error greater than $10^{-20}.$ This selection criteria can be adjusted through experimentation.

• Is there a better way to accomplish this task?

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Tuesday, May 24, 2022

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• Yesterday and today we worked on the cycloid and cardioid. The work is in the class files 20220523_A2P2.nb. You can find this notebook in our shared Dropbox directory \307_Files\2022\.

• While working on cycloid a student suggested an interesting variation on emphasizing motion of the wheel: to change the color of the wheel spokes as spokes touch the road. In class I was not able to implement this suggestion. After the class I came up with a function that does exactly what was suggested. It turns out that Floor[] and Mod[] can be adopted for this purpose. My solution is in 20220523_A2P2.nb. Of course, this is not the only way; it is an interesting challenge, you can look for your own solution.

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Friday, May 20, 2022 (updated)

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• 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 a function 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 specifiec 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

  

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Monday, May 16, 2022

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• 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 cardoid.
• 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 consturct 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.

    

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Tuesday, May 10, 2022

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• Today I summarized our explorations related to "funny" trigonometric functions. Over the weekend I posted "funny" trigonometric functions that parameterize regular $n$-gons. Notice that in the postings on my Mathematca file and in the postings over the weekend the code that is posted on the website can be copy-pasted into a Mathematica file, so that you do not need to retype it.
• All functions that I ask you to explore in Problem 1 are presented in the class files 20220505_A1P1.nb, 20220506_A1P1.nb and 20220509_A1P1.nb.
• Today we discussed recursively defined functions in Mathematica. There is a section about these functions at my Mathematica website. See also today's notebook 20220510_A1P2.nb.

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Sunday, May 8, 2022

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• I often look for new "funny" trigonometric functions. One idea that I explored today 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 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

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Saturday, May 7, 2022

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• Inspired by the parametrization of the unit circle by the functions $\cos$ and $\sin$ presented on Thursday, yesterday in class I started a derivation of a parametrization of a regular $n$-gon inscribed in the unit circle. Here $n$ 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.

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Thursday, May 5, 2022

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• The explorations that we did in class today are in the notebook 20220505_A1P1.nb. You can find this notebook in our shared Dropbox directory \307_Files\2022\.
• In the notebook TheBeautyOfTrigonometry_12.nb 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, taken as the coordinates of a moving point, they do 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. This is a good source for "funny" trigonometric functions.

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Thursday, March 24, 2022 (updated)

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• The Mathematica part of the class will start on Monday, May 2, 2022 in BH 215 at 2:00 pm.
• Before May 2nd, I will send you an email with the instructions how to set-up a Dropbox folder for your homework assignments and how to gain access to a shared Dropbox folder with the class materials.
• 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.

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