# Snowflake question

1. Jun 14, 2006

### nofx

I have a question to do with Koch's snowflake and anti snowflake..
I have no clue how to find the formula for the total area of each.
This is my table for snowflake

& this is it for anti-snowflake

I've realised that the area for both are finite, with the snowflake being 81x(8/5) and antisnowflake being 81x(2/5)

2. Jun 14, 2006

### eok20

since you know how much area is added, you can set up a recurrence relation. for the snowflake, A_n = A_n-1 + 27*(4/9)^(n-1). since A_n-1 = An-2 + 27*(4/9)^(n-2) and so forth, A_n = A_0 + 27*(4/9)^(1-1) + 27*(4/9)^(2-1) + ... + 27*(4/9)^(n-1). this is a geometric series with the ratio 4/9<1 so it will converge as n goes to infinity.

Last edited: Jun 14, 2006