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Author Topic: Proving that Pascal's triangle has a Fractal Structure  (Read 1681 times)
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PhillyWilliams
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« on: September 14, 2011, 05:38:53 PM »

I mentioned this on another post, but if you make Pascal's triangle all mod 2, you end up with Sierpinski's triangle, which is pretty surprising and amazing, but.... Is there a way to prove that Pascal's triangle must have a fractal structure much like Sierpinski's?

My thinking is that there's an intractable hurdle with this problem: the most common way of thinking about/constructing the standard Sierpinski triangle is by starting with an equilateral triangle, then iterating "down" towards the ultimate fractal, whereas Pascal's triangle starts with a single number and works "up" towards the ultimate fractal.  The two approaches, in some sense, are on opposite sides of infinity, which definitely causes problems.

I guess my question is: is there a line between "noticing" that a pattern exists and actually proving that the pattern must exist?  If no, how do we cross that line?
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fractower
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Posts: 173


« Reply #1 on: September 14, 2011, 09:27:08 PM »

I think I have a proof.

Instead of starting with a single one, start with a 1 in the middle of a line of zeros and consider a finite response to the isolated 1.

0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0
 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0

Notice how on the 4th step we end up with 4 ones and on the 5th step we end up with two ones seperatd by 4-1 zeros. Because these ones are isolated they will not interact for an additional 3 steps. So the response of these 1s will exactly match initial 1. At step 8 we will have 8 and at step 9 we are back to having 2 isolated 1s seperated by 8-1 zeros. Repeating the previous logic, the response of these ones will exactly match the response of the starting 1.

The final step is to apply induction to the 2^Nth step.
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fractower
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Posts: 173


« Reply #2 on: September 14, 2011, 09:38:05 PM »

OOPS. I think I just proved what you already know.
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PhillyWilliams
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« Reply #3 on: September 21, 2011, 02:17:42 PM »

(sorry for taking forever to respond -- with school started, there've been lots of distractors)

Yes, I knew that, but I think that's definitely a great way to prove it!  Considering the Sierpinski triangle created by Pascal's as a construction of ``smaller" triangles gets directly to the self-similarity of the fractal -- in other words, depending on what definition for a fractal you use, you can definitely satisfy a major condition that way.  Granted, it's not till you zoom WAY out that Pascal's triangle starts to technically become a fractal, but the goal here was to prove that its structure is fractal, which your proof, by aiming at self-similarity, hits the nail on the head. 

I think....

Thanks!
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fractower
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Posts: 173


« Reply #4 on: September 21, 2011, 03:12:50 PM »

Most fractals we are use to thinking about have a finite size but infinite microscopic detail. The Sierpinski you found in Pascal's triangle has finite microscope detail but infinite size. I think that means that one has infinitely more fractal cred or they are equal.  cheesy
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