Fractal Properties
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Just as the images above weren't very good pictures of a point, line, plane, or space, the drawing meant to be the Sierpinski Triangle has limitations. Remember as we continue that fractals are really formed by infinitely many steps. So there are infinitely many smaller and smaller triangles inside the figure, and infinitely many holes (the black triangles).
Let's organize our information into a table.
Let's look further at what we mean by dimension. Take a selfsimilar figure like a line segment, and double its length. Doubling the length gives two copies of the original segment.
Take another selfsimilar figure, this time a square 1 unit by1 unit. Now multiply the length and width by 2. How many copies of the original size square do you get? Doubling the sides gives four copies.Take a 1 by 1 by 1 cube and double its length, width, and height. How many copies of the original size cube do you get? Doubling the side gives eight copies.
Figure Dimension No. of Copies Line segment 1 2 = 2^{1} Square 2 4 = 2^{2} Cube 3 8 = 2^{3} Do you see a pattern? It appears that the dimension is the exponent  and it is! So when we double the sides and get a similar figure, we write the number of copies as a power of 2 and the exponent will be the dimension.
Let's add that as a row to the table.
Figure Dimension No. of Copies Line Segment 1 2 = 2^{1} Square 2 4 = 2^{2} Cube 3 8 = 2^{3} Doubling Similarity d n = 2^{d}
We can use this to figure out the dimension of the Sierpinski Triangle because when you double the length of the sides, you get another Sierpinski Triangle similar to the first.
Start with a Sierpinski triangle of 1inch sides. Double the length of the sides. Now how many copies of the original triangle do you have? Remember that the black triangles are holes, so we can't count them.
Doubling the sides gives us three copies, so
3 = 2^{d}, whered = the dimension.But wait, 2 = 2^{1}, and 4 = 2^{2}, so what number could this be? It has to be somewhere between 1 and 2, right? Let's add this to our table.
Figure Dimension No. of Copies Line Segment 1 2 = 2^{1} Sierpinski's Triangle ? 3 = 2^{?} Square 2 4 = 2^{2} Cube 3 8 = 2^{3} Doubling Similarity d n = 2^{d} So the dimension of Sierpinski's Triangle is between 1 and 2. Do you think you could find a better answer? Use a calculator with an exponent key (the key usually looks like this ^ ). Use 2 as a base and experiment with different exponents between 1 and 2 to see how close you can come. For example, try 1.1. Type 2^1.1 and you get 2.143547. I'll bet you can get closer to 3 than that. Try 2^1.2 and you get 2.2974. That's closer to 3, but you can do better.
Okay, I got you started; now find the exponent that gets you closest to 3, and that's its dimension. Email me your answer if you want.
lanius@rice.edu That's how fractals can have fractional dimension.
Note 2: Is there anything special about us doubling the lengths here? Could you have tripled them and derived the formula as well? Why don't you investigate by trying it.
Note 3: Robert Devaney has more information on fractal dimension.
Note 4: You may obtain a print version of this page.
Copyright 19972007 Cynthia LaniusURL http://math.rice.edu/~lanius/fractals/dim.html