Frozen Pages/Koch Snowflake

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Sol-Koch
Fields: Fractals and Dynamic Systems
Image Created By: SolKoll
Website: common.wikimedia.org

Sol-Koch

The image is an example of a Koch Snowflake, a fractal that first appeared in a paper by Swede Niels Fabian Helge von Koch in 1904. It is made by the infinite iteration of the Koch curve.


Contents

Basic Description

Curve Construction


The curve begins as a line segment and is divided into three equal parts. A equilateral triangle is then built, using the middle section of the line as its base, and the middle section is removed.

First 7 iterations
First 7 iterations


The Koch Snowflake is created by repeating the process of the Koch Curve on the three sides of an equilateral triangle an infinite amount of times in a process referred to as iteration. However, as seen with the animation, a complex snowflake can be created with only seven iterations - this is due to the butterfly effect of iterative processes.


As one might guess, a Koch Snowflake does have a finite area with can be calculated to be 8/5 of the original triangle, but the snowflake has an infinite perimeter that increases by 4/3 of the previous perimeter for each iteration.

A More Mathematical Explanation

Note: understanding of this explanation requires: *Calculus

Infinite Perimeter of the Koch Snowflake
Infinite Perimeter of the Koch Snowflake

The infinite perimeter of a Koch Snowflake can be explained by the fact that it is an example of an iterated process. To make the Koch Snowflake, each side of the Koch Snowflake infinitely undergoes the Koch curve process. Thus, each iteration produces additional sides that in turn produce additional sides in subsequent iterations. With a ever-increasing number of sides, the perimeter of the Koch snowflake will infinitely lengthen.

Furthermore, as seen in the animation, the Koch Snowflake displays a property known as self-similarity. This means that as we continue to magnify the Koch Snowflake, each magnified section continues to look similar to the larger perspective. However, there are still equations that can be used to determine various properties of this fractal.

Click here, for more information about Iterated Functions.

Number of Sides

The total number of sides on a Koch Snowflake with a degree of iteration (k) is given by N_k = (3)4^k\,, since the snowflake begins as a triangle with 3 sides, and each side turns into a Koch curve with 4 sides after each iteration.


Side Length

In addition, each time the snowflake undergoes an iteration, every side of the fractal is split into an equilateral triangle. Thus, the length of a side at any given degree of iteration can be determined by L_k = (x)3^{-k}\,, where x is the side length of the original triangle.


Perimeter

To calculated the perimeter of the fractal at any given degree of iteration, we multiply the number of sides by the length of each side:

P_k = N_k * L_k\,

P_k =(3)4^k * (x)3^{-k} = (3x)\left( \frac{4}{3} \right)^k = \left (\frac{4}{3} \right)^kP_0\,, where P_0 is the perimeter of the original triangle.

Thus, the perimeter increases infinitely by a ratio of 4/3 of the perimeter for the previous iteration.


Area

First iteration of the Koch Snowflake
First iteration of the Koch Snowflake
Second iteration of the Koch Snowflake
Second iteration of the Koch Snowflake
Third iteration of the Koch Snowflake
Third iteration of the Koch Snowflake

It is clear that the first iteration shown here adds to the area of the fractal: (N_1) A_0\frac{1}{9} = (3) \frac{1}{9}A_0\,

Now, to determine the finite area (A_\infty\,) of a Koch Snowflake, we assume the area of the original triangle is: A_0\,.

For the second iteration we add: (N_2)A_0\frac{1}{9^2} = (3)(4) \frac{1}{9^2}A_0\,

For the third iteration we add: (N_3) A_0\frac{1}{9^3} = (3)(4^2)\frac{1}{9^3}A_0\,


Thus the total area of the Koch Snowflake at the third iteration will be the summation of the expressions above: A_3 = A_0 + (3)\frac{1}{9}A_0 + (3)(4)\frac{1}{9^2}A_0 + (3)(4^2)\frac{1}{9^3}A_0 \,.


Or more generally:

A_k = A_0 \left(1 + (3)(4^0)\frac{1}{9^1} + (3)(4^1)\frac{1}{9^2} + (3)(4^2)\frac{1}{9^3} +...+ (3)(4^{k-1})\frac{1}{9^k} \right) \,.

A_k = A_0 \left (1 + \frac{3}{9}\left[1 + \frac{4}{9} + \left(\frac{4}{9}\right)^2 + \left(\frac{4}{9}\right)^3 + ... + \left(\frac{4}{9}\right)^k \right] \right)\,

A_k = A_0 \left [1 + \frac{3}{9}\sum_{k=0}^\infty \left(\frac{4}{9}\right)^k \right],which contains a geometric series that will converge


\lim_{k\rightarrow \infty}A_k= A_0\left[1+\frac{3}{9}\left(\frac{1}{5/9}\right)\right] = \frac{8}{5}A_0\,

Thus, the Koch Snowflake is approximately 8/5's of the area of the original triangle.


Fractal Dimension

2nd iteration of Koch Snowflake
2nd iteration of Koch Snowflake

We can also calculate the fractal dimension of a Koch Snowflake quite simply. For more information on Fractal Dimension.


Taking the image shown to the left, the top diagram shows that the new new Koch Curve lengths are a third of the previous iteration's length after the second iteration, and so e = 3. The bottom diagram shows that there are now a total of 4 Koch Curves, so that N = 4.


Thus, using the formula for fractal dimension: \frac{logN}{loge} = \frac{log4}{log3} \approx 1.26\,.




About the Creator of this Image

SolKoll has created various fractals, drawn using an iterated function system (IFS).







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