Fractals

March 16, 2023

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In 1904, Swedish mathematician Niels Fabian Helge von Koch discovered a curve with finite area but infinite length. Later, one would call these fractals. We start with an equilateral triangle. On each edge, we find the middle third. On the outside of that middle third, we build an equilateral triangle, and then remove the middle third. One can think about those steps as a single action: one takes any line segment and replaces its middle third with the remaining two edges of a triangle. We can do this action to all of the edges, and then do it again on all of the smaller edges, and so on. The end result of infinitely many of these actions is called the Koch (see the Figure below).

What is the area inside? What is the length of the snowflake? Let’s look at the length first. Each action replaced an edge with 4 edges, each \frac{1}{3} as long as before. Those 4 segments in total are \frac{4}{3} as long as the original edge. So, in each step, one is multiplying the total length by \frac{4}{3}. If one keeps doing that infinitely many times, then we are multiplying by \frac{4}{3}^{n}. And that goes to infinity since 4^{n} grows much faster than 3^{n}. We have solved that the Koch snowflake has an infinite perimeter. 

It turns out that the area must be finite – it is alway enclosed in a box. If we do the computation, each triangle is \frac{1}{9} the area of the previous one. The original triangle plus the sum of the new triangles is an infinite series.

We simplify this, and can use the formula for geometric series.

The area of the Koch snowflake is the area of the original triangle times \frac{8}{5}. The entire figure is just 60% more than the original, but the perimeter is infinite. Thus, it has a finite area, but the perimeter is infinite. Note that there is no contradiction here. The area and perimeter are just different; they are two different attributes of a figure. 

References:

Lee, C. & Sheppard, J. (2020). Fractals and the Koch snowflake. University of Leicester.

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