Calculus Project: Fractals and Infinite Series

John Bidwell, David Brenner, and Ben Geiger

The Koch Snowflake

The Koch curve first appeared in Swedish mathematician Helge von Koch's 1904 paper entitled "Sur une courbe continue sans tangente, obtenue par une construction géométrique élémentaire." To form the curve, first divide a line segment into three equal segments. Next, draw an equilateral triangle using the middle line segment as the base. Then, remove the base of this new triangle. Finally, repeat this process over and over again. The Koch curve is the limit approached as this process is repeated. The Koch Snowflake is the same as the Koch curve, only beginning with an equilateral triangle instead of a single line segment.

The significance of the Koch curve is that it has an infinite perimeter that encloses a finite area. To prove this, the formulas for the area and the perimeter must be found. To find these formulas, however, the formulas for the number of sides and the length of each side must first be found. To do so, first draw the original triangle and examine the first few iterations. On the first triangle, there are three sides. The second triangle has 12 sides. The third triangle has 48 sides. From these observations, the general formula for the number of sides can be found. Likewise, examine each iteration for the length of each side. Notice that the length is reduced by a factor of one-third at each iteration. From this, the general formula for the length of a side at each iteration can be found.

Let x = length of original side.

Let n=0 denote the first iteration.

The number of sides , sn (at the nth iteration), is equal to sn = 3 · 4 n.

The length of a side , ln (at the nth iteration), is equal to ln = 

x
3 n
.


Now, it is possible to derive the formulas for the perimeter and the area of the snowflake. Perimeter is the total length of the sides of the snowflake. This would be equal to the number of sides multiplied by the length of each side. To prove that this perimeter is infinite, take the limit as n approaches infinity.

The perimeter, pn (at the nth iteration), is equal to pn =

x · 3 · 4 n 
3 n 
.

The

lim
x · 3 · 4 n
3 n
n → ∞

 = ∞ because 4n increases much faster than 3n.

Because it is greater than 1, it fails the ratio test and therefore diverges.

From these formulas it is also possible to derive a formula for the area of the snowflake, as well as to prove that this area is finite even though the perimeter is infinite. To derive this formula, it is easiest to draw out a few iterations of the snowflake and write out the series to determine the general form. The area, An, of the snowflake will be the original area of the triangle plus the sum of the area of each new triangle multiplied by the number of new triangles. To prove the area is finite, take the limit as n approaches infinity of An.

Area =

1
2
  · b · h.

The area of the original triangle =

1
2
  · 
\3
2
  · x 2.


base = ln.

height = ln · 

\3
2
.

The area of each triangle is equal to (ln)2 · 

\3
4

The number of triangles is equal to 3 · 4 n-1 starting at n=1.

The area, An (at the nth iteration), is equal to

 
An
 
 
 = 
 
\3
4
        x 2
 
 
 + 
 
n­
Σ
\3
4
        (ln) 2· 3 · 4 n-1
k = 1­
   
.

 
limAn
n → ∞

 
 
 = 
 
\3
4
    x 2
 
 
 + 
 
­
Σ
\3
4
    (ln) 2· 3 · 4 n-1
k = 1­
.

 
limAn
n → ∞

 
 
 = 
 
\3
4
    x 2
 
 
 + 
 
­
Σ
\3
4
    · 
x 2
9 n
    · 3 · 4 n-1
k = 1­

.

 
limAn
n → ∞

 
 
 = 
 
\3
4
  x 2
 
 
 + 
 
­
Σ
\3
16
  x 2 
n
 
4
9
­­
 
k = 1­
.

 
limAn
n → ∞

 
 
 = 
 
\3
4
    x 2
 
 
 + 
 
 
 x2 
 
\3
12
1 − 
4
9
.

limAn
n → ∞
  = 
\3
5
  x 2.

Below is a flash document that will display the first five iterations of the Koch Snowflake and the area, perimeter, number of sides, and length of each side at each iteration.

 

The Anti Snowflake

The anti snowflake is extremely similar to the Koch snowflake. It is formed by dividing the edges of an equilateral triangle into three parts, removing the middle piece, and adding a section that creates an indentation in the triangle. The anti snowflake, like the Koch snowflake, has an infinite perimeter with a finite area. The formulas for the number of sides, the length of the sides, and the perimeter are the same, however the area formula changes. Instead of adding the area of the new triangles formed, the area of these "new triangles" is subtracted.

Let x = length of original side.

Let n=0 denote the first iteration.

The number of sides , sn (at the nth iteration), is equal to sn = 3 · 4 n .

The length of a side , ln (at the nth iteration), is equal to ln = 

x
3 n
.

The perimeter, pn (at the nth iteration), is equal to pn =

x · 3 · 4 n 
3 n 
.

The

lim
x · 3 · 4 n
3 n
n → ∞
 = ∞ because 4n increases much faster than 3n. (See the proof by the ratio test above).

 

The area, An (at the nth iteration), is equal to

 
An
 
 
 = 
 
\3
4
                x 2
 
 
 - 
 
n­
Σ
\3
4
                (ln) 2· 3 · 4 n-1
k = 1­
   
.

The limit as n approaches infinity of the area of the anti snowflake is equal to:

 
limAn
n → ∞

 
 
 = 
 
\3
4
    x 2
 
 
 - 
 
 
 x2 
 
\3
12
1 − 
4
9
.

limAn
n → ∞
  = 
 
\3
10
  x 2.

 

Project Contributions

 

John Bidwell: Arranged meeting times/places, created the java program.

David Brenner: Created the web site and flash document, wrote the report.

Ben Geiger: Wrote the formulas, quality control for both web site and java program.

 

Java Program

 

Unfortunately, we were unable to transfer the program into a java applet to run on the web site. However, the code for the program as well as pre-compiled classes are available at the links below. They can be compiled and should run fine on the user's individual machine.

KochSim.java KochSim.class

Triangle.java Triangle.class

AntiTriangle.java AntiTriangle.class