Exploration 9: Fractals

Exploration 9: Fractals
A fractal is a geometric shape that commonly exhibits the property of selfsimilarity. A self-similar object is one whose component parts resemble the
whole. Another way to think of this is that each part of the object, when magnified, looks much
like the object as a whole. We will start looking at this property with the Koch Snowflake. This
fractal, pictured above, was first described in 1904 by the Swedish mathematician Helge von
Koch.
The Koch Snowflake
You can think of the Koch Snowflake as being composed of three similar sides. We will focus on
just one of these sides and build it up through stages creating what is known as a Koch Curve.
We start with a line segment as shown in Figure 1. We call this step 0. At each step, the middle
third of every line segment is replaced by a triangular “roof.” The length of each segment of the
roof is a third of the length of the previous line segment.
Start or Step 0
Step 1
Step 2
Figure 1: The first two steps in making Koch Curve.
Question 1: Continue what was done in Figure 1, by drawing steps 3 and 4 of a Koch Curve.
Suppose you kept completing step after step in making a Koch Curve and did this indefinitely. If
you then looked at say the left third of this structure and magnified it three times, it would look
exactly the same as your Koch Curve. This is means that it has the property of self-similarity.
The figure shown at the very beginning of this section is a stage 3 Koch Snowflake. It is compose
of three of the sides that you created in Question 1. As the number of steps increase, the visual
changes you can see in the figure are less and less. Figure 2 shows a stage 4 Koch Snowflake.
This should look quite similar to the one shown earlier. Imagine what a stage 6, or stage 100, or
a stage 1000 Koch Snowflake would look like. Unless you could magnify it a great deal, they
would all look very similar to that shown in Figure 2. If you could, however, continue this
process indefinitely, we would have a true Koch Snowflake and it would be a true fractal.
Figure 2: A stage 4 Koch Snowflake. Not a true fractal yet, but starting to look like one.
Question 2: As we moved from step to step in making the Koch Curve in Question 1, we
increased the number of sides. Complete the following table by finding the number of
segments at step 3, step 4, and step n of a Koch Curve.
0
1
Step
Number of Segments
1
4
2
16
3
4
n
Question 3: Assuming the segment at step 0 of making a Koch Curve has length 1, complete
the following table by finding the sums of the lengths of the segments at steps 2, 3, 4, and n.
As n approaches infinity, what does the sum of the lengths of the segments approach?
Step
Sum of the Segment Lengths
0
1
1
4∙
2
16 ∙ 3
4
n
Hopefully, from Question 3, you should realize that a Koch Curve has a length of infinity. If this
is true, then total length of all three sides (or the perimeter) of a Koch Snowflake is infinite in
length. This is the amazing part. Even though its perimeter is infinite in length its area is not.
For example, our level 4 Koch Snowflake is shown in Figure 3. If it were to become a snowflake
of level 5, level 6, and so on, its perimeter would get larger and larger, but its area will never
increase past the circle that we placed it in.
Figure 3: A Koch Snowflake has a perimeter that is infinite in length, but has finite area.
The Sierpinski Triangle
Another interesting fractal is that of the Sierpinski Triangle. It was first described by Waclaw
Sierpinski, a polish mathematician, in 1915. To construct this, we start with a filled-in triangle
and then remove the triangle whose vertices are the midpoints of the sides of the triangle. We
show this in Figure 4. For each solid triangle remaining, the “upside-down middle” triangle is
removed in the second step. This process continues indefinitely.
Stage 0
Stage 1
Stage 2
Figure 4: The first couple of steps in creating a Sierpinski Triangle.
Question 4: Continue what was done in Figure 4 by drawing stage 3 and stage 4 Sierpinski
Triangles.
Question 5: Let’s count the number of triangles at each stage of a Sierpinski Triangle. Do
this by completing the following table.
Stage
Number of Triangles
0
1
1
3
2
9
3
4
n
Question 6: Now let’s take a look at the area of shaded regions in a Sierpinski Triangle. We
will assume our initial triangle has area 1. Complete the following table to find the areas of
the different stage triangles. As n approaches infinity, what does the area of the shaded
region approach?
Stage
Area of Shaded Region
0
1
1
3/4
2
3
4
n
From Question 6, you should be able to see that if we continued to construct our Sierpinski
triangle indefinitely, it would have area zero. A rather remarkable concept! What will happen
to the length of the boundary of this figure?
Question 7: Suppose the perimeter of our original equilateral triangle shown in Figure 4 is
one. Complete the following table to determine the boundary of the Sierpinski Triangle at
various stages. As n approaches infinity, what does the boundary of the shaded region
approach?
Stage
Boundary of Shaded Region
0
1
1
3/2
2
3
4
n
From Question 7, you should have seen that a completed Sierpinski triangle would have a
boundary of infinity. Pretty amazing!
Figure 5 shows the first three iterations of the Sierpinski Carpet. For this fractal we start with a
square. In the first stage, the square is subdivided into nine smaller squares and the interior
square is removed. In the second stage, each of the remaining eight squares is subdivided into
nine squares and the interior squares are removed. This process is repeated indefinitely to
make the Sierpinski Carpet.
Stage 0
Stage 1
Stage 3
Stage 2
Figure 5: The beginning of the Sierpinski Carpet.
Question 8: Assume the initial square used to construct a Sierpinski Carpet has area one.
Complete the following table to find the areas of the shaded regions for the different stage
carpets. As n approaches infinity, what does the area of the shaded region approach?
Stage
Area of Shaded Region
0
1
1
2
3
4
n
Question 9: Figure 6 shows the beginning stages of the Sierpinski Tetrahedron. Construct a
stage 2 Sierpinski Tetrahedron using paper and tape. Carefully cut out a net for a
tetrahedron, fold it into a tetrahedron and tape it. Make four of these and then tape them
together as shown in Figure 6 to make a stage 1 Sierpinski Tetrahedron. Complete four
stage 1 Sierpinski Tetrahedrons to make a stage 2 Sierpinski Tetrahedron. (Tetrahedron
nets can be found in Template A at the end of this exploration. There are also directions at
the end of the exploration on how to construct tetrahedrons out of business cards.)
Stage 0
Stage 1
Stage 2
Figure 6: The first couple of stages for a Sierpinski Tetrahedron.
The Dragon Curve
The next fractal we want to look at is that of the dragon curve. It is named as such because its
shape resembles that of a dragon. Various iterations of these can be found on the pages of the
book Jurassic Park. Figure 7 shows a drawing of the 14th iteration of a dragon curve. Can you
see a dragon in the drawing? Notice that it appears to be composed of a series of similar, but
different size “islands.”
Figure 7: The 14th iteration of a dragon curve.
One of the ways to begin to construct a dragon curve is done with simple paper folding. Start
with a long narrow piece of paper. Construction paper works well because it is slightly stiff. Cut
a strip that is about an inch wide (or about 3 cm) and as long as your sheet of paper. Fold it over
once by putting the right edge of your long strip over to the left edge. Make a crease and then
open it up so that a right angle is formed by the fold. This is the first iteration of the dragon
curve. (See Figure 8.) Fold your paper back to how it was before you opened it. Now fold your
paper again by putting the right edge over to the left edge. Make another crease and again
open it up so that all angles that are formed are right. You now have the second iteration of a
dragon curve. Now you can just repeat this process over and over.
Figure 8: The first three iterations of a dragon curve.
Question 10: Using paper folding, make the fourth and fifth iterations of a dragon curve.
Make a sketch of your results.
You probably found that making the fifth iteration of the dragon curve is just about as much as
you can by folding paper. It would be quite difficult, if not impossible, for you to fold the paper
in half many more times. We can, however, continue to make more iterations in a drawing if we
just analyze our results first and figure out a pattern. After all, we are just dealing with a path of
line segments that turn left or right at right angles. (See Figure 9.)
L
L
R
1st Iteration
R
R
2nd Iteration
R
L
R
L
R
R
3rd Iteration
Figure 9: The first three iterations of the dragon curve labeled with right or left.
Question 11: Complete the following table by including the directions taken when traveling
down the path of the 4th and 5th iterations of the dragon curve.
Iteration
1st
2nd
3rd
4th
5th
Directions Right or Left
R
RRL
RRLRRLL
Question 12: There is a recursive pattern to your lists of Rs and Ls from Question 11.
Determine what the pattern is and write down the string of Rs and Ls for the 6th iteration of
the dragon curve. As a hint, Figure 10 shows a picture of the 3rd iteration where the paper is
not quite completely folded out. Make a drawing of the 6th iteration of the dragon curve.
R
L
R
L
R
L
R
Figure 10
Dimension
It is fairly intuitive that a line has one dimension, a square has two dimensions, and a cube has
three dimensions. How about more complicated shapes like fractals? How many dimensions do
they have? Let’s start by looking at the simple objects and define how we calculate dimension.
In Figure 11, we show a line segment, a square, and a cube that have each been scaled up by a
factor of two. Some people would call this making it “twice as large.”
2 copies
4 copies
8 copies
Figure 11: Scaling up a line segment, square and a cube by a factor of two.
To make a line segment “twice as large,” we would obviously have to make it twice as long.
How do we make a square “twice as large?” If you put a square in a copy machine and set the
scaling to make it 200% of the original (or “twice as large”) it will print out a new square that is
twice as wide and twice as high. It would take four of the original size squares to make the same
size square as our larger one. What if we have a cube and make it “twice as large?” That is to
say we want to make all sides of the cube twice as long. We would then need eight of our
smaller cubes to make one of our larger cubes.
Because each our three objects had different dimensions it took a different number of them to
make a new one that is scaled by a factor of two. (Perhaps “twice as large” is not a good phrase
for this since scaling a square by a factor of two really makes it four times as large.)
•
•
•
It took 2 (or 21) line segments to scale it by a factor of 2.
It took 4 (or 22) squares to scale it by a factor of 2.
It took 8 (or 23) cubes to scale it by a factor of 2.
Hopefully you can see that the exponent on the 2 in our three examples is the same number as
the dimension of our objects.
Let’s see if this works for scaling an object up by a factor of three. How many copies of a square
are needed to scale it up by a factor of three? Hopefully you can see that it will take 9 of them.
(See Figure 12.) Since 9 = 32 and a square has dimension 2, this seems to work.
2
3 = 9 copies
Figure 12: It takes nine copies of a square to scale it up by a factor of three.
Let’s put this idea into a formula. If d is the dimension of an object and n is the number of
copies needed to scale an object by a factor of s, then the following relationship is true.
= n
Let’s see how this works for some of our fractals. In Figure 13, we have two pictures of a Koch
Curve. The one on the right is three times as long and three times as high as the one on the left.
In other words, it is scaled up by a factor of three. How many copies of the smaller version,
does it take to make the larger?
Figure 13: A Koch Curve and one scaled up by a factor of three.
If our Koch Curve were a one-dimensional object, it would take 31 = 3 copies to scale it up by a
factor of three. If this were a two-dimensional object it should take 32 = 9 copies. However, as
you can see, it takes 4 copies of the smaller to produce the larger. Since 4 is in between 3 and 9,
this fractal has a dimension that is between 1 and 2. Kind of freaky! As you can see, objects do
not need to have dimensions that are nice whole numbers. Let’s approximate the dimension of
our Koch Curve. To do that we need to find the solution to
3 = 4.
To solve this we will need the help of logarithms. We first take the log of both sides of our
equation, use a logarithm property you may have seen before, and then divide as follows.
3 =4
log 3 = log 4
∙ log 3 = log 4
log 4
=
log 3
Using our calculator we see that log 3 / log 4 ≈ 1.26. This means that the Koch Curve has a
dimension of about 1.26.
We don’t need to go through all the algebra we just did to find dimensions of other fractals. If
we just solve = n for d, we can have a nice formula that will more easily give us our answer. If
we solve = n for d using logarithms as we did earlier we get
=
log
log
where d is the fractal dimension, n is the number of smaller fractals needed to cover the larger
fractal and s is the scale factor between the smaller and larger fractals.
Question 13: With the aid of Figure 14, determine the dimension of the Sierpinski Triangle.
Figure 14
Question 14: A Sierpinski Carpet is shown in Figure 15. Determine its dimension.
Figure 15
Question 15: A hexaflake is shown in Figure 16. Determine its dimension.
Figure 16
Question 16: Determine the dimension of a Sierpinski Tetrahedron. Use Figure 6 as a guide.
It is a bit more difficult to calculate the dimension of the dragon curve; however its result is
simple. It has dimension 2. Another nice property of the dragon curve is that it can be used to
tile a plane. (See Figure 17.)
Figure 17
Template A: Tetrahedron nets.
Constructing a Tetrahedron from Business Cards
To make a single tetrahedron you will need two standard U.S business cards (2” by 3.5”). Fold a
card so that the bottom right corner touches the upper left corner.
Fold the flaps over the equilateral triangles that you just created and make your creases sharp
so it makes an impression in the paper. Now open up the card slightly. This is a "left-handed
unit."
We now repeat this same process with another card except that you start by folding the card so
that bottom left-hand corner touches the upper right-hand corner. This will be a "right-handed
unit."
Slip the two units together so the flaps hook on the inside
and sort of hold the two cards together. You will then need
to secure it with a couple small pieces of tape. After you
make a number of these tetrahedrons you can put them
together, securing with tape, to make a Sierpinski
Tetrahedron.
Exercises
1. A stage 0 Koch curve consists of just a straight line segment. Draw a stage 2 Koch curve.
2. Draw three stage 3 Koch curves so it forms a Koch snowflake.
3. Assume the initial square used to construct a Sierpinski Carpet (as shown below) has area
one and the perimeter is four.
Stage 0
Stage 1
Stage 3
Stage 2
a) Complete the following table to find the areas and perimeters of the shaded regions for
the different stage carpets.
Stage
Area of Shaded Region
0
1
Perimeter of Shaded Region
4
1
2
3
n
b) As n approaches infinity, what does the area of the shaded region approach?
c) As n approaches infinity, what does the perimeter of the shaded region approach?
4. A stage 0 Sierpinski Tetrahedron consists of a single tetrahedron. What is the stage of the
Sierpinski Tetrahedron shown in the picture?
5. In describing the dragon curve, we used R for a right turn and L for a left turn. Using this
notation, the third iteration was RRLRRLL. What is the fourth iteration?
6. What is the dimension of the Sierpinski Hexagon shown below?