Snowflake By Haese Mathematics -

Using the sum of a geometric series with ratio ( r = \frac{4}{9} < 1 ), as ( n \to \infty ): [ A_{\infty} = A_0 \left[ 1 + \frac{1}{3} \times \frac{1}{1 - \frac{4}{9}} \right] = A_0 \left[ 1 + \frac{1}{3} \times \frac{9}{5} \right] = A_0 \left[ 1 + \frac{3}{5} \right] = \frac{8}{5} A_0 ]

For each side, remove the middle third and replace it with two segments of the same length (forming an equilateral "bump"). The number of sides increases. snowflake by haese mathematics

Introduction One of the most captivating paradoxes in mathematics is the Koch Snowflake, a fractal curve first described by Swedish mathematician Helge von Koch in 1904. In the context of Haese Mathematics (particularly for IB Diploma Analysis & Approaches SL/HL), the snowflake serves as a perfect case study for geometric sequences , limits to infinity , and the surprising behaviour of infinite series . Construction: The Iterative Process We begin with an equilateral triangle of side length ( a_0 ). This is Stage 0. Using the sum of a geometric series with

Repeat the process: take every straight line segment, divide it into three equal parts, and replace the middle third with two segments of that length. In the context of Haese Mathematics (particularly for

This simplifies to: [ A_n = A_0 \left[ 1 + \sum_{k=1}^n \frac{3 \times 4^{k-1}}{9^k} \right] = A_0 \left[ 1 + \frac{1}{3} \sum_{k=1}^n \left(\frac{4}{9}\right)^{k-1} \right] ]

Since ( A_0 = \frac{\sqrt{3}}{4} ), the final area is: [ A_{\infty} = \frac{8}{5} \cdot \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{5} ]