Is the Fractal Perimeter Infinite but Area Finite?

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SUMMARY

The discussion centers on the Koch Snowflake, a well-known fractal that exhibits an infinite perimeter while enclosing a finite area. Participants debate the implications of adding smaller segments with each iteration, suggesting that this could lead to convergence. However, it is established that the perimeter diverges due to the increasing number of segments added at each step. The conversation highlights the paradox of fractals and invites further exploration of their mathematical properties.

PREREQUISITES
  • Understanding of fractals, specifically the Koch Snowflake
  • Basic knowledge of infinite series and convergence
  • Familiarity with mathematical notation and summation
  • Experience with graphing tools like Graphamatica 2.0e
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  • Research the mathematical properties of the Koch Snowflake and its perimeter-area paradox
  • Explore the concept of convergence in infinite series, focusing on examples like the harmonic series
  • Study other fractals and their equations, such as the Sierpinski Triangle and Mandelbrot Set
  • Learn how to graph fractals using Graphamatica 2.0e or similar software
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Mathematicians, students studying calculus or geometry, and anyone interested in the properties of fractals and their applications in mathematical theory.

Gib Z
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Hey if you guys look up Fractal on Wikipedia, you see the author states that the Koch Snowflake, a common and famous fractal, supposedly has an infinite perimeter yet finite area. It sed it would be infinite perimeter because it keeps on adding perimeter with each iteration. How ever, i thought since it keeps on adding Less with each iteration, it would remsemble a series that continually adds less. i haven't worked out the actual series yet, i will soon, but basically since it keeps adding less and less, it should eventually converge into a finite number eventually, right? sort of like if u kept on adding 10^0 + 10^-1 + 10^-2 + 10^-3 + 10^-4 so on so forth, sure u keep adding numbers, but the first one is 1, 2nd term 0.1, 3rd 0.01, 4th in 0.001, so on in that fashion, adding to 1.11111111111111111111111..., or 1 and 1/9. well yea, so this fractal i think really had finite area
 
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Not all series which have terms that go to zero converge. For example,

\sum_{n=1}^\infty \frac{1}{n} = \infty

To see this, note:

1+ \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4} \right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} +\frac{1}{8} \right) + ... > 1 + \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4}\right) + \left( \frac{1}{8} +\frac{1}{8} +\frac{1}{8} +\frac{1}{8} \right) + ... = 1 + \frac{1}{2} +\frac{1}{2} + ... = \infty

But this doesn't matter here, because in the Koch snowflake, at each step you increase the perimeter by 4/3, so the terms in the series actually increase, and so it obviously diverges.
 
Last edited:
Gib Z said:
How ever, i thought since it keeps on adding Less with each iteration, it would remsemble a series that continually adds less.
It is true that each triangle part adds less per iteration but at the same time there are more of them per iteration. :smile:
 
yea that's for the help, i just realized wen i actually bothered to work it out this morning. wow that is quite extra ordinary, infinite perimeter inside a finite area. does anyone know the equations of any fractals, hopefully that i can plug into Graphamatica 2.0e?
 
Well, no, the perimeter is NOT "inside" the area!
 
excuse me? yes i think quite so. I was no referring to inside the finite area of the fractal, but say, a square around it.
 

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