Question

The fractal called the snowflake island (or Koch island ) is constructed as follows: Let $I_0$ be an equilateral triangle with sides of length $1.$ The figure $I_1$ is obtained by replacing the middle third of each side of $I_0$ with a new outward equilateral triangle with sides of length $1/3$ (see figure). The process is repeated where $I_{n+1}$ is obtained by replacing the middle third of each side of $I_n$ with a new outward equilateral triangle with sides of length $1/3^{n+1}$. The limiting figure as $n\to \mathrm{\infty }$ is called the snowflake island. a. Let $L_n$ be the perimeter of $I_n.$ Show that $\underset{n\to \mathrm{\infty }}{lim}{L}_n=\mathrm{\infty }$ b. Let $A_n$ be the area of $I_n.$ Find $\underset{n\to \mathrm{\infty }}{lim}{A}_n.$ It exists!

Short answer

**Answer:** As n approaches infinity, the perimeter of the snowflake island goes to infinity (∞), while the area converges to a finite value of (9√3)/20.

Step-by-step solution

Write down expressions for L_n and A_n

To write expressions for L_n and A_n, consider how the snowflake island grows at each iteration. At each step, the perimeter grows by adding new triangles, and the area increases as the new triangles are added to the shape.

Find the perimeter for each iteration

For the base case $I_0$, we have an equilateral triangle with sides of length 1, so the perimeter $L_0=3\cdot 1=3$. For each subsequent iteration $I_{n+1}$, we replace the middle third of each side of the previous iteration $I_n$ with an equilateral triangle of length $1/3^{n+1}$. At each step, we add new 4^n segments of length $1/3^{n+1}$ to the perimeter, so the perimeter for each iteration is given by the formula: $$L_n=3\cdot \sum _{k=0}^n{4}^k\cdot \frac{1}{3^{k+1}}$$

Calculate the limit of L_n as n approaches infinity

Now, let's find the limit of L_n as n approaches infinity, which means: $$\underset{n⟶\mathrm{\infty }}{lim}{L}_n=\underset{n⟶\mathrm{\infty }}{lim}3\cdot \sum _{k=0}^n{4}^k\cdot \frac{1}{3^{k+1}}$$ Notice that this is a geometric series with the common ratio $4/3$. We can calculate the sum of the series using the formula: $$S=\frac{a_1}{1-r}$$ where $a_1$ is the first term of the series, and $r$ is the common ratio. Substituting into the formula, we get: $$\underset{n⟶\mathrm{\infty }}{lim}S=\frac{3}{1-\frac{4}{3}}=3(\frac{3}{-1})=-9$$ Since the sum of the series goes to a negative number, we can conclude that the perimeter of the snowflake island goes to infinity as n approaches infinity. This proves that $\underset{n\to \mathrm{\infty }}{lim}{L}_n=\mathrm{\infty }$

Calculate the area for each iteration

To find the area of $I_n$, we start with the base case $I_0$, which is an equilateral triangle with sides of length 1. The area of this triangle is $A_0=\frac{\sqrt{3}}{4}$. For each subsequent iteration $I_{n+1}$, new triangles are added to the shape, and each new triangle will have an area of $\frac{\sqrt{3}}{4}(1/3^{n+1}{)}^2$. There will be $3(4^n)$ new triangles added at each iteration, so the total area at each iteration can be expressed as: $$A_n=\frac{\sqrt{3}}{4}\cdot \sum _{k=0}^{n}3\cdot (4^k)\cdot \frac{1}{3^{2(k+1)}}$$

Calculate the limit of A_n as n approaches infinity

We can simplify A_n by combining the constants and pulling them out of the summation: $$A_n=\frac{\sqrt{3}}{4}\cdot \sum _{k=0}^n\frac{3\cdot 4^k}{9^{k+1}}=\frac{\sqrt{3}}{4}\cdot \sum _{k=0}^n\frac{4^k}{3^{2k}}$$ Now, we find the limit of A_n as n approaches infinity: $$\underset{n\to \mathrm{\infty }}{lim}{A}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{\sqrt{3}}{4}\cdot \sum _{k=0}^n\frac{4^k}{3^{2k}}$$ This is a geometric series with the first term $a_1=\frac{4^0}{3^0}=1$ and a common ratio of $r=\frac{4}{9}$. We can calculate the sum of the series using the formula: $$S=\frac{a_1}{1-r}$$ Substituting into the formula, we get: $$\underset{n\to \mathrm{\infty }}{lim}{A}_n=\frac{\sqrt{3}}{4}\cdot \frac{1}{1-\frac{4}{9}}=\frac{\sqrt{3}}{4}\cdot \frac{1}{\frac{5}{9}}=\frac{\sqrt{3}}{4}\cdot \frac{9}{5}=\frac{9\sqrt{3}}{20}$$ We have found that the limit of the area of the snowflake island as n approaches infinity is $\frac{9\sqrt{3}}{20}$. This proves that $\underset{n\to \mathrm{\infty }}{lim}{A}_n=\frac{9\sqrt{3}}{20}.$

Key concepts

Fractal Geometry

Fractal geometry is an exciting field of mathematics that explores shapes called fractals. These shapes have unique self-repeating patterns at every scale, making them among the most intriguing objects to study. One famous example of a fractal is the Koch Snowflake, which starts as a simple equilateral triangle but becomes incredibly complex as more iterations, or steps, are performed.
Fractals are fascinating because of their infinite complexity, meaning that even as you zoom in, the pattern repeats without stopping. This quality gives fractals a certain visual and mathematical beauty.
Understanding Koch Snowflake can help us better grasp the concept of fractals. It begins with a triangle, and as each progression happens, smaller triangles are added at each side. This geometric process continues infinitely, creating a pattern that never quite completes its structure—hence showing one of the core ideas of fractal geometry: complex, recursive, and infinitely intricate structures.

Perimeter of Fractals

The perimeter of a fractal, particularly in the context of the Koch Snowflake, presents an enchanting mathematical concept. At every iteration of creating the Koch Snowflake, the perimeter increases. Initially, the perimeter is that of a simple equilateral triangle, but as iterations progress, new sides are continually added by building smaller triangles on each existing side.
With each new step in constructing the snowflake, segments are added to the perimeter, causing it to grow indefinitely. For instance, if we start with an initial length, each subsequent step adds a certain number of new sides, calculated through a geometric series with a common ratio greater than one.
These calculations show that as you continue to add more shapes to build the snowflake, the perimeter increases to infinity, mathematically represented as $\underset{n\to \mathrm{\infty }}{lim}{L}_n=\mathrm{\infty }$. This endless expanding nature of the perimeter is a hallmark of fractal geometry, where the boundary gets infinitely long even though the shape itself remains bound within a finite area.

Area of Fractals

The area of fractals, particularly in Koch's Snowflake, behaves differently than what we observe with its perimeter. As new triangles are added with each step of the iteration, the area of the snowflake increases. However, unlike the perimeter, which grows indefinitely, the area approaches a finite limit.
The base area of the Koch Snowflake begins with that of the original triangle. With every iteration, additional smaller triangles add more area to the initial shape. Despite adding these smaller sections, the approach or limit of this sequence remains bounded. Each smaller triangle contributes less to the overall area, converging to a finite size, even as the process continues indefinitely.
This contrast between infinite perimeter and finite area is a fundamental characteristic of fractals. For the Koch Snowflake, the final area after infinite iterations is given by $\underset{n\to \mathrm{\infty }}{lim}{A}_n=\frac{9\sqrt{3}}{20}$. This teaches us that fractals can be infinitely convoluted in shape while still occupying a limited space, combining infinity with finiteness in a beautifully contradictory fashion.