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Today we worked on Problem 2 on Assignment 3. The notes are in the notebook 20210604_A3P2.nb. As always, this notebook is in the folder 307_Files\2021 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:
{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 are Goldbach lengths of 500000 positive even integers, my recreation of the picture from Wikipedia.
- What I value about Problem 1 on Assignment 2 is that with rigorous analysis, we can understand the behavior of the function ff[x], which numerical calculations can never capture. Never! There will always be limitations to numerical calculations. There are also limitations to rigorous thinking. What is the moral of the story? In a quest for a genuine understanding of a mathematical problem, we must utilize all our resources: our capacity to think rigorously and the computational facilities that the technology is making available to us.
- To enforce the power of the synergy of the rigor and the technology, I ask: What would the joy of solving a square wheel problem algebraically be without seeing a square wheel move in an animated gif file? In engineering schools, they make it a student project to create a bicycle with square wheels with an accompanying road for a smooth ride. It must be a unique feel to experience the smooth ride on square wheels. But, in a math class, we stop at an animated gif. We enjoy the power that math gives to engineers to create physical objects.
- The synergy of rigor and technology is the recent crowning of the preceding millennia of human efforts.
- My first idea was to do the square wheel instead of the classical cycloid and cardioid on Assignment 2. However, after working out the animation of a square wheel myself, I thought that it might be too technical for the first problem involving mathematical motion. I want to share the animations that I created. So, if you need some entertainment and have Mathematica handy, you can work on these animations. The equations of arcs on which $n$-gon is rotating are upside down catenaries, that is, scaled and shifted arcs of the graph of a cosine hyperbolic reflected across the $x$-axes. The lengths of the chords over which the arcs are spanned are \[ 2 \cos\left(\dfrac{\pi}{n}\right) \operatorname{arccosh}\left(\dfrac{1}{\cos\left(\dfrac{\pi}{n}\right)}\right) \]
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Here $L= \sqrt{2} \operatorname{arccosh}\bigl(\sqrt{2}\bigr) \approx 1.2464505$
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Here $L= \phi \operatorname{arccosh}\left(\dfrac{2}{\phi}\right) \approx 1.0910006$, where $\phi = \dfrac{1+\sqrt{5}}{2}$ is the Golden Ratio
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Here $L= \sqrt{3} \operatorname{arccosh}\left(\dfrac{2}{\sqrt{3}}\right) \approx 0.95142615$
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Here $L= 2 \cos\left(\dfrac{\pi}{7}\right) \operatorname{arccosh}\left(\dfrac{1}{\cos\left(\dfrac{\pi}{7}\right)}\right) \approx 0.83730864$
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Here $L= \sqrt{2+\sqrt{2}} \operatorname{arccosh}\left(\dfrac{2}{\sqrt{2+\sqrt{2}}}\right) \approx 0.74501594$
- Today I started coding of the Koch snowflake in the file 20210528_A3P3.nb which is in 307_Files\2021. In the past I constructed the Koch curve in vonKoch_curve.nb in the same directory. In the same directory there is a more complicated fractal called Pythagora's Tree: PythagorasTree.nb, see also the corresponding pdf printout PythagorasTree.pdf in the same directory. All these files can serve as "guides" for constructions of fractals in Problem 3.
- In the first part of Problem 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.
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Today we studied a geometric construction related to Problem 2 on Assignment 2. We construucted the involute of the unit circle. Let us celebrate this construction by the animation below.
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- In Problem 2 on Assignment 2 you are asked to explore two famous curves a cycloid and a cardioid. First we discuss a cycloid.
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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:
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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.
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.
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In the first part of Problem 2 I ask you to reproduce the following animations.
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- In the second part of Problem 2 I ask you to explore a cardoid.
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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*}
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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.
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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.
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Below is the cardioid generated by a wheel with radius 1/2 rolling on a circle with radius 1.
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Below is the cardioid generated by a wheel with radius 1/3 rolling on a circle with radius 1.
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Below is the cardioid generated by a wheel with radius 2 rolling on a circle with radius 1.
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Below is the cardioid generated by a wheel with radius 3/2 rolling on a circle with radius 1.
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Below is the cardioid generated by a wheel with radius 1/2 rolling on a circle with radius 1.
- The Problem 1 on Assignment 1 is inspired by the trigonometric functions Cos[t] and Sin[t]. 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.
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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.
- 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. In class, we discussed 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.
- We discussed Problem 1 on Assignment 1 today. The notebook 20210506_A1P1.nb contains this discussion. A typical definition of a function in Mathematica is given in the snippet posted below. First Clear[ ] the name of the functions that you will use and the variables that you will use. The syntax of a function definition is that the variables in the definition on the left-hand side have underscores. Separate Clear[ ] and the definitions with semicolon ;.
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Inspired by the parametrization of the unit circle presented above, today I derived a parametrization of a regular $n$-gon inscribed in the unit circle. Here $n$ is a positive integer greater than $2.$ In class I derived the functions analogous to Cos[t] and Sin[t].
Here is a sketch of that derivation which I did in class. We derived the parametrization functions to be
- 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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- The Mathematica part of the class will starts on Monday, May 3, 2021 as an online class at 2:00 pm. You will get a Canvas invitation for the Zoom meetings.
- In the meantime I will send you an email with the instructions how to set-up a Dropbox folder for your homework assignments, how to gain access to a shared Dropbox folder with the class materials and how to gain remote access to BH 215 computer lab computers.
- 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.