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Topic: Uh-oh spaghettios (math)
»Waisztarroz«
Pancake
posted 12-01-2002 02:04:52 PM
My math teacher gave our class something similar to Sierpinski's Triangle as the non-routeine problem for this week (it isn't labeled that, but it starts out in the first diagram as one triangle, then three triangles with one upside down in the middle of them (think triforce), then the figure that would follow that). What I have to find is how many "segments" are in the nth figure (3 in the first diagram). If you blank out all the upside-down triangles, you get 3*the number of triangles for the number of segments. So, I've been trying to come up with a flat out equation for the number of right-side up triangles in the nth diagram, but that's probably not possible with a fractal, is it?

I'm glad this one is extra credit, heh.

Although I suppose it's not exactly a fractal, it tends to act like one, I think.

[ 12-01-2002: Message edited by: »Waisztarroz« ]

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Xelloss Metallium
Pancake
posted 12-01-2002 02:12:06 PM
*stares off in space and blinks* Umm.. I think you need to go ask.. Someone who even knows what you are talking about And besides. I think by saying that you are confusing everyone into utter madness
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»Waisztarroz«
Pancake
posted 12-01-2002 02:17:38 PM
Uhh, nevermind, I figured out an equation for the number of segments in figure n.

So, I'm set.

I guess I was kinda wrong in thinking it was a fractal even though it looked like one. It has 1 triangle (3 segments) in the first figure, 3 triangles (9 segments) in the second figure, 10 triangles (30 segments) in the third figure, and 36 triangles (108 segments) in the fourth figure. With the way it forms a bunch of little "triforces" over and over it reminded me of sierpinski's triangle...

[ 12-01-2002: Message edited by: »Waisztarroz« ]

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»Waisztarroz«
Pancake
posted 12-01-2002 03:10:44 PM
And, just to help reduce any insanity the thread caused, here's the page.

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»Waisztarroz«
Pancake
posted 12-01-2002 04:16:30 PM
I see why 2. says "Explain how" instead of "What is?" since the result (of 3*(1*(2^(100-2))+2*(4^(100-2))), the number of segments) is over 10^59.

Still, if I had the capability, I'd find all 59 digits and write them down.

[ 12-01-2002: Message edited by: »Waisztarroz« ]

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Razor
posted 12-01-2002 04:23:00 PM
Using the Maple Math Engine in Scientic Workplace it is

6.02601766597121353328235784628886713895997294719669840642048 x 10^59

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»Waisztarroz«
Pancake
posted 12-01-2002 09:26:17 PM
How cool is this?
quote:
[20:04] <Waisztarroz> !math 3*(1*(2^98)+2*(4^98))
[20:04] <ChanServ> 602601766597121353328235784628886713895997294719669840642048
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