Koch's Snowflake 2

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, which is made by the infinite iteration of the Koch curve.

Basic Description

The Koch Snowflake was first described in 1904, but it was not until 1967 that some of its most significant properties became apparent. That year, Benoit Mandlebrot published a paper explaining an earlier finding that as the length of a country's border is measured with increasingly fine measurement devices, the measured length does not approach a specific, "true" measurement of the border, as one would expect with increasingly precise measurement. Instead, the measured length grows exponentially, suggesting that given a fine enough measuring device, there would be no upper limit on how long the measured length of the border could be.

Mandlebrot's paper showed for the first time that this strange finding could be explained if the border of a country resembled a fractal, shapes which are largely defined by self-similarity No matter how much you zoom in on one small section of a fractal, parts of it continue to look like the whole original border, with just as much complexity. , and have since been found throughout nature in such places as coastlines and the branching of blood vessels.

Mandlebrot's paper went on to show that this self-similarity could lead to other counter-intuitive properties in fractals as well. For instance, Mandlebrot showed that it doesn't make sense to think of most fractals as being 1 or 2 dimensional, but rather as something in between.

To derive and demonstrate how self-similar shapes can behave in these ways, Mandlebrot didn't actually use coastlines, but rather used variations of the Koch Snowlake, because of its relative geometric simplicity. With this example, he was able to show how these surprising properties could exist in any fractal and to provide insights into how they might be measured and interpreted. These insights continue to form the basis of much of our analysis of fractals today.

Koch Curve Construction

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

First 7 iterations

The Koch Snowflake is an iterated process. It 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). Thus, each iteration produces additional sides that in turn produce additional sides in subsequent iterations.

An interesting observation to note about this fractal is that although the snowflake has an ever-increasing number of sides, its perimeter lengthens infinitely while its area is finite. The Koch Snowflake has perimeter that increases by 4/3 of the previous perimeter for each iteration and an area that is 8/5 of the original triangle.

Click here, for more information about Iterated Functions.

A More Mathematical Explanation

Note: understanding of this explanation requires: *Calculus in some sections

Infinite Perimeter of the Koch Snowflake

Fractal Properties

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Self-similarity

The Koch Snowflake displays a property known as self-similarity, emphasized in the animation. This means that as we continue to magnify the Koch Snowflake, each magnified section continues to look similar to the larger perspective.

Fractal Dimension

2nd iteration of Koch Snowflake

The fractal dimension of a Koch Snowflake can be calculated using the formula for fractal dimension: $\frac{logN}{loge}$.

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.

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

Demonstration

Other Properties

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  Another Koch Snowflake

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  Number of Sides

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  Side Lengths

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  Perimeter

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  Area

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Number of Sides ($N_k\,$) [show more][hide]

Number of Sides

Let us observe the number of sides of a Koch Snowflake at its first iterations:

The starting triangle has a total number of sides: $3 = 3*1 = 3*4^0\,$

The 1st iteration has a total number of sides: $12 = 3*4 = 3*4^1\,$

The 2nd iteration has a total number of sides: $48 = 3*16 = 3*4^2\,$

With the first three iterations, we can see a pattern emerging. The "3" coming from the sides of the starting triangle, and the "4" reflecting the number of new sides that emerge from each side of the previous iteration.

Thus, the total number of sides on a Koch Snowflake with a degree of iteration (k) is given by $N_k = (3)4^k\,$.

Side Length ($L_k\,$) [show more][hide]

Side Lengths

The length of the sides of the Koch Snowflake at each iteration also follows a pattern. Each time the snowflake undergoes an iteration, every side of the fractal creates a protruding equilateral triangle so that the side lengths shorten by a factor of 3.

Each side of the starting triangle has a length: $x \,$

Each side of the 1st iteration has a length: $\frac{x}{3} = \frac{x}{3^1} \,$

Each side of the 2nd iteration has a length: $\frac{x}{9} = \frac{x}{3^2} \,$

The length of a side at any given degree of iteration (k) can be determined by $L_k = (x)3^{-k}\,$, where x is the side length of each of the three sides of the original triangle.

Perimeter ($P_k\,$) [show more][hide]

Perimeter

As stated before, the perimeter of a Koch Snowflake lengthens infinitely. Let us observe the first few iterations of the snowflake to determine by what ratio the perimeter increases at each iteration. 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:

$P_k = N_k * L_k\,$

$P_k =(3)4^k * (x)3^{-k} = \left( \frac{4}{3} \right)^k(3x) = \left (\frac{4}{3} \right)^kP_0\,$, 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.

As the number of iterations approaches infinity, we can see that the perimeter also approaches infinity: $\lim_{k\rightarrow \infty}P_k= \left (\frac{4}{3} \right)^kP_0 = \infty$.

Area ($A_k\,$) [show more][hide]

First iteration of the Koch Snowflake

Second iteration of the Koch Snowflake

Third iteration of the Koch Snowflake

Finally, let us examine the area of the Koch Snowflake and prove that it is indeed finite. We assume that the area of the original triangle is $A_0\,$, and determine the area that each of the first few iteration adds to the total previous area (see the images to the right for visual illustrations).

The original area is: $A_0\,$

The 1st iteration adds: $(3) \frac{1}{9}A_0 = \frac{(3)}{9}A_0\,$

The 2nd iteration adds: $(12)\frac{1}{9^2}A_0 = \frac{(3)(4)}{9^2}A_0\,$

The 3rd iteration adds: $(48)\frac{1}{9^3}A_0 = \frac{(3)(4^2)}{9^3}A_0\,$

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

Or more generally, to find the area ($A_k$) at a degree of iteration k:

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

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

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

$A_k = A_0 \left [1 + \frac{3}{9}\sum_{k=0}^\infty \left(\frac{4}{9}\right)^k \right]$, which contains a geometric series that will converge $\frac{1}{1-4/9} = \frac{1}{5/9}$

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

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

Teaching Materials

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References

Cynthia Lanius, Cynthia Lanius' Fractal Unit:Koch Snowflake Larry Riddle, Koch Curve

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