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Homework Help: Snowflake question

  1. Jun 14, 2006 #1
    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
    http://img485.imageshack.us/img485/8959/snowflake2an.jpg [Broken]
    & this is it for anti-snowflake
    http://img356.imageshack.us/img356/2616/antisnowflake8vz.jpg [Broken]

    I've realised that the area for both are finite, with the snowflake being 81x(8/5) and antisnowflake being 81x(2/5)
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jun 14, 2006 #2
    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
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