Today something weird happened.
We’ve already seen the Sierpinski Triangle appear in two ways, by iterating the process of breaking a triangle into four equal pieces and removing the central one, and by coloring the even elements of Pascal’s triangle. Today we found a third way, by Iterating a Curve.
By starting with a solid, filled in triangle, and step-by-step removing stuff from that area, we reach the same exact figure as we get by starting with a simple line segment and lengthening/bending it. The area clearly starts as 2-dimensional, the line segment clearly starts as 1-dimensional. Yet both processes have the same end result. What does this mean for the dimension of Sierpinski’s Triangle? Is it 2D or 1D? It certainly can’t be both, so the only option is… neither?
When I first took this course (back in the days of Mr. Drix), this was the first moment where I realized all of this talk about fractional dimension may be more than nonsense. Maybe there’s something to it after all…
Your homework: think critically about what dimension really means. What does it mean to say something is 1-dimensional or 2-dimensional. We’ll dig into this in the next few days…
We finished our discussion of the Sierpinski triangle, noting that just like the “final” version of the Koch Curve is “nothing but angles,” this geometric oddity is “nothing but edges,” as the area of the triangle converges to zero as the iterations continue. This conclusion also presented an interesting contradiction. For the Koch curve, we argued that an infinite number of segments, each of length zero, resulted in an infinite perimeter (effectively, ∞ * 0 = ∞). Here, we have an infinity of triangles, each again with an area of zero, resulting in an area of zero (effectively, ∞ * 0 = 0).
What this reveals is that the expression “∞ * 0” is what is called an Indeterminate Form, an expression the defies definition. We can create a reasonable argument that defines it as infinity, and we can create a just-as-reasonable argument that defines it as zero. Therefore, it must be defintionless.
We finished the day by opening up the PC laptop mobile lab and downloading FractaSketch to each device. We’ll be using this software extensively over the next few weeks!
We finished our discussion about the Cantor Set, noting that infinite set of endpoints that are left over with each segment removal are countably infinite. If the claim is that the cantor set is actually uncountable, that requires there to be other elements of the set that are not segment endpoints. And there are such points, uncountably infinitely many of them. See this post for more information on how this works.
We went on to draw a picture of the Koch Snowflake, a figure devised by Swedish mathematician Helge von Koch as an example of a continuous curve with no tangents. As the fractalization process continues, the number of segments that make its perimeter increases without bound, but the length of each segment shrinks to zero. Mathematically, however, the total perimeter also increases, resulting in a figure with finite area and infinite perimeter!
We ended the period by starting to draw an image of another fractal, called the Sierpinski Triangle. Start with an equilateral triangle (connect the dots in the worksheet), then bisect and connect all three sides. Fill in that middle triangle, effectively removing it and leaving you with three triangles at the three corners. Then do that process again for each of the three triangles: bisect the sides, connect them, and fill in the middle triangle. Do that as many steps as you can fit. Don’t cheat and look up what the final result looks like!