Koch Snowflake - Part 1

While this a question related to common sense ; still I have a Theoretical Explanation ( by the help of a for loop used in C++ ) :

( 1 ) The Area : In a Triangle,the recursive progress of area :

for( i = 1; i > length ; i++ ) {

Side = side1 + side2 +side3; Recursive = Side/3;

Recursive1 = Recursive - side1; Recursive2 = Recursive - side2; Recursive3 = Recursive - side3;

RFinal = Recursive1 * Recursive2 * Recursive3; RArea = Rfinal*Recursive ;

Area = sqrt(RArea) ;

cout << Area; }

This code will return a finite value.

(ii)Perimeter :

as the perimeter increases infinitely, let l,m,n be the sides of the triangle.

for( i = 1; i > l ; i++) { for( i = 1; i > m ; i++) { for( i = 1; i > n ; i++) { cout<<l + m + n; } } }

This loop will run Infinitely.

Therefore the answer is : a Triangle will have finite area and infinite perimeter for Koch Snowflake curve.

The Snowflake has a finite area because every time it grows , it always encloses a finite area. But as it grows indefinitely its perimeter grows continuously hence has an infinite perimeter.

Call the side length, number of sides, area, and perimeter of the fractal to be $l_i, n_i, a_i, p_i$ at time step $i$ , letting $l_1 = 1$ . Also let $a = \dfrac{\sqrt{3}}{4}$ , the area of a single unit triangle, so $a_1 = a$ . Note that $p_i = l_in_i$ , and trivially $n_1 = 3$ .

Every time step, the side length is divided by $3$ , and the number of sides is multiplied by $4$ . Thus $l_{i+1} = \dfrac{l_i}{3}$ and $n_{i+1} = 4n_i$ , so $p_{i+1} = l_{i+1}n_{i+1} = \dfrac{l_i}{3} \cdot 4n_i = \frac{4}{3} l_in_i = \frac{4}{3} p_i$ . Thus $p_i$ is a geometric sequence with ratio greater than $1$ , and thus it diverges to infinity: it has infinite perimeter .

Recall $l_{i+1} = \dfrac{l_i}{3}$ and $n_{i+1} = 4n_i$ . Together with $l_1 = 1, n_1 = 3$ , we have $l_i = \dfrac{1}{3^{i-1}}$ and $n_i = 3 \cdot 4^{i-1}$ .

Also, on time step $i$ to $i+1$ , we add $n_i$ tiny triangles of side length $l_{i+1}$ . Call the area added on time step $i$ to be $b_i$ ; thus $a_i = \sum_{j=1}^i b_i$ . Thus $b_i$ is:

$n_i \cdot \dfrac{l_{i+1}^2 \sqrt{3}}{4}$

$= n_i l_{i+1}^2 a$

$= 3 \cdot 4^{i-1} \cdot \left( \dfrac{1}{3^i} \right)^2 a$

$= \dfrac{4^{i-1}}{3^{2i-1}} a$

$= \frac{3}{4} a \cdot \left( \frac{4}{9} \right)^i$

and so by induction $b_{i+1} = \frac{4}{9} b_i$ for $i \ge 2$ . Thus $b_i$ is a geometric sequence with ratio between $0$ and $1$ . As the series $b_i$ converges, and $a_i$ is just the partial sums $b_i$ , we must have $a_\infty$ to be finite, or finite area .

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Also, just in case you don't want to get your hands dirty, just head to Wikipedia and get the answer.

For an area, just draw a square surrounded the snowflake and you'll know the truth.

If you put a box around the snowflake, you could enclose the snowflake for all steps. But, there are an infinite number of tiny triangles on the snowflake, making for an infinite perimeter.

isnt it obvious , for any polygon for that instance that recursively progresses will have finite area and infinite perimeter .