## Archive for December, 2007

### The Starry Night

December 31, 2007

Vincent van Gogh‘s magnum opus, eternal The Starry Night (1889):

A higher definition image can be found here. The original painting can be found at the MoMA in New York City.

### London Pianist

December 28, 2007

[ photo courtesy of Alexandra Z ]

### Perimeter of the Koch Snowflake

December 28, 2007

A couple of days ago, I wrote a post about the Koch snowflake and how to compute its area. However, the perimeter of the Koch snowflake is (in my humble opinion) more interesting than its area.

As I wrote in my previous post, we draw the Koch snowflake by starting with an equilateral triangle, and then following some simple rules ad infinitum. At each iteration $k \geq 0$ we will have a snowflake, a regular polygon of $n_k$ sides, each of length $l_k$. As we have seen before, at each iteration, the number of sides of the snowflake is

$n_k = 4^k 3$

and the length of each side is given by

$l_k = 3^{-k} x$.

The perimeter of the snowflake is simply

$P_k = n_k l_k = \left(\displaystyle\frac{4}{3}\right)^k P_0$,

where $P_0 = 3 x$ is the perimeter of the initial snowflake (i.e., the equilateral triangle whose side is $x$). Note that the perimeter of the snowflake is given by a geometric progression of common ratio $r = 4/3$ (which is $> 1$), and therefore the progression $\{P_k\}_{k=0}^{\infty}$ will diverge: the perimeter of the snowflake will tend to infinity as $k \rightarrow +\infty$. The Koch snowflake has a finite area, but a boundary of infinite length! Wow! Ain’t that cool!?

Note also that the boundary of the Koch snowflake is continuous but not differentiable anywhere.

### Area of the Koch Snowflake

December 26, 2007

A Koch Snowflake is a fractal curve that can be constructed quite easily. Start with an equilateral triangle, and then iterate as follows:

1. Divide each line segment into three segments of equal length.
2. Draw an equilateral triangle that has the middle segment from step 1 as its base and points outward.
3. Remove the line segment that is the base of the triangle from step 2.

The following animated image illustrates the first seven iterations of this algorithm:

[ source ]

What is the area of the Koch snowflake at each iteration $k \in \mathbb{N}$? I have solved this problem years ago, when I was in high school. Nonetheless, a post on the Koch snowflake at Jeffrey Wiens‘ blog made me want to solve the problem yet once again.

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My solution:

I will use the following variables:

• $A_k$: area of the snowflake.
• $n_k$: number of sides of the snowflake.
• $l_k$: length of the sides of the snowflake.

where the subscript denotes iteration $k \in \mathbb{N}$. Let us also define function $a(s) = \frac{\sqrt{3}}{4} s^2$, which gives us the area of the equilateral triangle whose side is $s > 0$. The variables above are initialized as follows:

$A_0 = a(x), \quad n_0 = 3, \quad{} l_0 = x$

since that the initial snowflake is an equilateral triangle with side $x$. It can easily be seen that the length of each side is reduced by a factor of $3$ with each iteration, and therefore

$l_{k+1}= 3^{-1} l_k$.

Since $l_0 = x$, we have $l_k = 3^{-k} x$. It can also be seen that the number of sides is increased by a factor of $4$ with each iteration, and therefore

$n_{k+1} = 4 n_k$.

Since $n_0 = 3$, we have $n_k = 4^k 3$. The area of the snowflake at each iteration is given by the following recursion

$A_{k+1} = A_k + n_k a(l_{k+1})$.

We know that $A_0 = a(x)$, and we can now compute the area at each iteration:

$A_1 = A_0 + n_0 a(l_1) = a(x) + 3 a(x/3)$

$A_2 = A_1 + n_1 a(l_2) = a(x) + 3 a(x/3) + 4.3 a(x/3^2)$

$A_3 = A_2 + n_2 a(l_3) = a(x) + 3 a(x/3) + 4.3 a(x/3^2) + 4^2 3 a(x/3^3)$

and, finally, we obtain

$A_k = a(x) + \displaystyle\frac{3}{4} \displaystyle\sum_{m=1}^k 4^m a(x/3^m)$.

Note that

$a(x/3^m) = a(3^{-m}x) = 3^{-2 m} a(x) = 9^{-m} a(x)$

and therefore we can compute the area of the snowflake at each iteration

$A_k = a(x) \left[1 + \displaystyle\frac{3}{4} \displaystyle\sum_{m=1}^k (4/9)^m \right]$.

We have a sum of $k$ terms of a geometric progression with common ratio $r = \frac{4}{9}$

$\displaystyle\sum_{m=1}^k r^m = r \displaystyle\sum_{m=0}^{k-1} r^m = r \left[\frac{1-r^k}{1-r}\right] = \left(\frac{r}{1-r}\right)(1 - r^k)$,

and therefore

$\displaystyle\sum_{m=1}^k (4/9)^m = \left(\frac{4/9}{1-4/9}\right)(1 - (4/9)^k) = \frac{4}{5} (1 - (4/9)^k)$

and finally we have the area of the snowflake at iteration $k$

$A_k = a(x) \left[\displaystyle\frac{8}{5} - \displaystyle\frac{3}{5} \left(\frac{4}{9}\right)^k\right]$.

Note that the Koch snowflake is a fractal curve, not a polygon. Therefore, the Koch snowflake is the curve to which the snowflake “converges” when we apply the recursion above ad eternum. Note also that for finite $k$ we have a polygon of $n_k$ sides, not a fractal curve. Having said that, the area of the Koch snowflake is thus given by

$\displaystyle\lim_{k \rightarrow +\infty} A_k = \displaystyle\frac{8}{5} a(x)$.

Since $a(x) = \frac{\sqrt{3}}{4} x^2$, we have

$\displaystyle\lim_{k \rightarrow +\infty} A_k = \displaystyle\frac{8}{5} \frac{\sqrt{3}}{4} x^2 = \displaystyle\frac{2 \sqrt{3}}{5}x^2$.

The area of a Koch snowflake is thus finite if $x$ is finite.

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