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.
Archive for December, 2007
The Starry Night
December 31, 2007London Pianist
December 28, 2007
[ photo courtesy of Alexandra Z ]
Perimeter of the Koch Snowflake
December 28, 2007A 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 
and the length of each side is given by
The perimeter of the snowflake is simply
where 
Note also that the boundary of the Koch snowflake is continuous but not differentiable anywhere.
Area of the Koch Snowflake
December 26, 2007A Koch Snowflake is a fractal curve that can be constructed quite easily. Start with an equilateral triangle, and then iterate as follows:
- Divide each line segment into three segments of equal length.
- Draw an equilateral triangle that has the middle segment from step 1 as its base and points outward.
- 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 
__________
My solution:
I will use the following variables:
: area of the snowflake.
where the subscript denotes iteration 
since that the initial snowflake is an equilateral triangle with side 
Since 
Since 
We know that 
and, finally, we obtain
Note that
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]](http://s0.wp.com/latex.php?latex=A_k+%3D+a%28x%29+%5Cleft%5B1+%2B+%5Cdisplaystyle%5Cfrac%7B3%7D%7B4%7D+%5Cdisplaystyle%5Csum_%7Bm%3D1%7D%5Ek+%284%2F9%29%5Em+%5Cright%5D&bg=ffffff&fg=333333&s=0 "A_k = a(x) \left[1 + \displaystyle\frac{3}{4} \displaystyle\sum_{m=1}^k (4/9)^m \right]")
We have a sum of 
![\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)](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Csum_%7Bm%3D1%7D%5Ek+r%5Em+%3D+r+%5Cdisplaystyle%5Csum_%7Bm%3D0%7D%5E%7Bk-1%7D+r%5Em+%3D+r+%5Cleft%5B%5Cfrac%7B1-r%5Ek%7D%7B1-r%7D%5Cright%5D+%3D+%5Cleft%28%5Cfrac%7Br%7D%7B1-r%7D%5Cright%29%281+-+r%5Ek%29&bg=ffffff&fg=333333&s=0 "\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
and finally we have the area of the snowflake at iteration
![A_k = a(x) \left[\displaystyle\frac{8}{5} - \displaystyle\frac{3}{5} \left(\frac{4}{9}\right)^k\right]](http://s0.wp.com/latex.php?latex=A_k+%3D+a%28x%29+%5Cleft%5B%5Cdisplaystyle%5Cfrac%7B8%7D%7B5%7D+-+%5Cdisplaystyle%5Cfrac%7B3%7D%7B5%7D+%5Cleft%28%5Cfrac%7B4%7D%7B9%7D%5Cright%29%5Ek%5Cright%5D&bg=ffffff&fg=333333&s=0 "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
Since 
The area of a Koch snowflake is thus finite if
__________
Some links:
