No matter how far down we recurse, the shape will never grow outside this hexagon, so it can't keep growing forever! As shown above, the Koch snowflake is self-similar with six pieces scaled by 1/3 and one piece scaled by \(1/\sqrt{3}\). THE KOCH SNOWFLAKE DRAWING SNOWFLAKES We can draw a certain snowflake by following these steps. The Koch snowflake along with six copies scaled by \(1/\sqrt 3\) and rotated by 30° can be used to tile the plane [].The length of the boundary of S(n) at the nth iteration of the construction is \(3{\left( {\frac{4}{3}} \right)^n} s\), where s denotes the length of each side of the original equilateral triangle. How do we go about calculating the area at every depth? This is a little more complicated to calculate.First let's consider what happens to the number of sides.When we first start out, there are 3 sides to the triangle, each of length one unit.On the next iteration, there are 12 sides, each of length 1/3 unit (Each of the three straight sides of triangle is replaced with four new segments).
This gives the value for the Area of the snowflake with an infinite depth. It's possible to continuously zoom into a fractal and experience the same behavior.The Koch curve is named after the Swedish mathematician Here is an animation showing the effect of zooming in to a Koch curve.The typical way to generate fractals is with recursion. This recursive procedure is continued and in the limit will produce the image of a space filling curve shown above, known as the flowsnake (a play on words of snowflake). The value for area asymptotes to the value below.If you look closely at the formulae you will see that the limit area of a Koch snowflake is exactly 8/5 of the area of the initial triangle.Below is a graph showing how the area of the snowflake changes with increasing fractal depth, and how the length of the curve increases. A Koch snowflake has a finite area, but an infinite perimeter! Each of the following iterations adds a number of triangles 4 times the previous one.
The area is therefore \(\frac{\sqrt{3}}{4}\). It is based on the Koch curve, which appeared in a 1904 paper titled "On a Continuous Curve Without Tangents, Constructible from Elementary Geometry" by the Swedish mathematician Helge von Koch. This produces the following attractor consisting of 6 smaller copies of the snowflake scaled by 1/3 and one copy in the middle scaled by \(1/\sqrt{3}\). This variation on the Koch snowflake was created by William Gosper. The boundary of the snowflake consists of three copies of the
The first iteration adds 3 triangles. A Koch snowflake has a finite area, but an infinite perimeter!Koch snowflakes of different sizes can be tesellated to make interesting patterns:Here's an interesting relationship between Koch curves and Thue-Morse sequences.To first understand the relationship, we first need to understand a Thue-Morse sequence!The ThueâMorse sequence (or ProuhetâThueâMorse sequence), is an infinite binary sequence obtained by starting with 0 and successively appending the Boolean complement of the sequence obtained thus far.If you are familiar with the educational programming language Logo, and Turtle graphics, it's possible to turn these sequences of binary digits into Koch curves.I think you can see where this is going. The snowflake area asymptotes pretty quickly, and the curve length increases unbounded. The area of each of these new triangles is added to the total area:Below is the equation showing how the area of the snowflake increases with each depth. Let's keep with the notation that the length of the side of initial triangle is The area of the first iteration is simply the area of the base triangleFor the next iteration the area of the snowflake is increased by the three red triangles shown in the diagram below(Three new triangles with sides of length s/3):For the next iteration we add an additional 12 smaller triangles. Click on the link above for more details. Even though we know the length of the all the line segments is increasing with each step, it's looking like the area is not getting that much bigger with successive terms? Other interesting facts. The boundary can be constructed by the following L-system: Let's consider what happens with each iteration (and we only need to consider one side; to get the total, we simply multiply by three).Each time we step down, the length of each side is replaced by What about the area? Notice that this will place the copy of the motif inside the hexagon containing that edge. If we imagine that each of these four line segments are, themselves, made up from smaller versions of themselves, the curve starts to form …The curve gets ever increasingly longer, more convulated, and 'twisty', even though the geometric distance between the end points remains the same. The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below). The red line segments are oriented by the arrows.
If so, what is this value? )Let's start with the easy part.
A fractal is a self-similar shape.Fractals are never-ending infinitely complex shapes. In order to find the sum, it helps if we clean this up a little.Then we can pull out any additional 1/4 from the bracket (in order to multiply each term inside by four):Using the following equality (flipping the exponents), we can move the square to the denominator and the increasing power to the numerator.Now all the exponents are the same order and we can combine them and pull the 4 inside each of the terms. If you zoom into a fractal, you get see a shape similar to that seen at a higher level (albeit it at smaller scale). Here is one example of an iterated function system for the Koch snowflake. Start with a pattern of seven regular hexagons. Here is another iterated function system for the Koch snowflake.