Since geeky science questions seem to be today's fashion...
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Q: Describe a 2 or 3 dimensional shape with an infinite edge and zero area, which takes up a finite amount of space.
Asymptote. y=1/x.
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Asymptote. y=1/x.
you mean a hyperbole (which BTW has two asymptotes). "takes up a finite amount of space" is debatable now, you need a lot of space to store one without folding, bending or cutting it. :)
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Good friggin call. Though I wouldn't say it has an infinite edge. Perhaps an infinite number of edges, but each of them of a finite length. And what happens when the cumulative edge length approaches infinity? Doesn't it approach having a surface? Who knew there was so much philosophy in such a simple question? :rolleyes:
well, either edge is used for perimeter, not the line connecting two vertices; or we could agree the vertices are all melting together and all edges become one... the surface issue is the problematic one, they are fractals after all. :)
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Q: Describe a 2 or 3 dimensional shape with an infinite edge and zero area, which takes up a finite amount of space.
I'm surprised that no one has suggested fractals. The one I had in mind is called the Sierpinski Triangle. Observe: 1. Draw an equilateral triangle. 2. For each triangle, find the mid-point of its sides and draw lines connecting them, creating four triangles. 3. Remove the triangle in the middle, leaving three equilateral triangles connected at their vertices, each having three edges that are half of the starting triangle. 4. Go to step 2. The triangle starts with three edges of length x, so its total perimeter is 3x. After the first iteration, the shape has nine edges -- three on each of the three triangles -- each of which have a length of x/2, meaning the shape's total perimeter is 9x/2, longer than what we started with. After the first iteration, the area is 3/4 what it was before. As you continue with more iterations, the number of edges increases without bounds, and so does the shape's perimeter. As the number of iterations approaches infinity, so do the number of edges and, consequently, the length of its perimeter. Also, each iteration decreases the area geometrically: as the number of iterations approaches infinity, the area bound by the perimeter approaches zero. The shape itself never exceeds the bounds set by the starting triangle, which makes it finite. With a variant called the Sierpinski Carpet, you start with a single square, divide it into nine squares, remove the center one and repeat. Again, the number of edges and the perimeter approach infinity while the area bound by the perimeter approaches zero. There are also 3-d versions of these shapes, called sponges, which start with a tetrahedron and a cube respectively. Added: Oops, a bit of a screw up. The number of edges is doubled, not tripled; I was counting the starting edges of the triangle twice. The increase in the length of the perimeter is still 9x/2, however, as you have the three starting edges (x + x + x = 3x or 6x/2) plus the three edges of the now empty center triangle (x/2 + x/2 + x/2 = 3x/2)
modified on Thursday, May 6, 2010 8:31 PM
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you mean a hyperbole (which BTW has two asymptotes). "takes up a finite amount of space" is debatable now, you need a lot of space to store one without folding, bending or cutting it. :)
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Luc Pattyn wrote:
you mean a hyperbole
NOW who's using hyperbole. ;P
Luc Pattyn wrote:
takes up a finite amount of space
Depends on how you define space. If you consider bounding rectangle, it takes up infinite space. If you consider the bound between the equation and the axes, I'm pretty sure that area is finite (if I felt like polishing my calculus skills, I could probably calculate exactly how much that area is). If you consider the amount of volume the curve itself displaces, it would take up no space at all. :)
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I'm surprised that no one has suggested fractals. The one I had in mind is called the Sierpinski Triangle. Observe: 1. Draw an equilateral triangle. 2. For each triangle, find the mid-point of its sides and draw lines connecting them, creating four triangles. 3. Remove the triangle in the middle, leaving three equilateral triangles connected at their vertices, each having three edges that are half of the starting triangle. 4. Go to step 2. The triangle starts with three edges of length x, so its total perimeter is 3x. After the first iteration, the shape has nine edges -- three on each of the three triangles -- each of which have a length of x/2, meaning the shape's total perimeter is 9x/2, longer than what we started with. After the first iteration, the area is 3/4 what it was before. As you continue with more iterations, the number of edges increases without bounds, and so does the shape's perimeter. As the number of iterations approaches infinity, so do the number of edges and, consequently, the length of its perimeter. Also, each iteration decreases the area geometrically: as the number of iterations approaches infinity, the area bound by the perimeter approaches zero. The shape itself never exceeds the bounds set by the starting triangle, which makes it finite. With a variant called the Sierpinski Carpet, you start with a single square, divide it into nine squares, remove the center one and repeat. Again, the number of edges and the perimeter approach infinity while the area bound by the perimeter approaches zero. There are also 3-d versions of these shapes, called sponges, which start with a tetrahedron and a cube respectively. Added: Oops, a bit of a screw up. The number of edges is doubled, not tripled; I was counting the starting edges of the triangle twice. The increase in the length of the perimeter is still 9x/2, however, as you have the three starting edges (x + x + x = 3x or 6x/2) plus the three edges of the now empty center triangle (x/2 + x/2 + x/2 = 3x/2)
modified on Thursday, May 6, 2010 8:31 PM
Gregory.Gadow wrote:
Remove the triangle in the middle
Ohhhhhhh.
Gregory.Gadow wrote:
I'm surprised that no one has suggested fractals.
Actually, Luc did. Though I was missing the point above that you mentioned.
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That would be a fractal, such as this Sierpinski triangle[^]. ADDED Although not many would agree they have 2 or 3 (or any integer) number of dimensions... :)
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modified on Thursday, May 6, 2010 6:06 PM
Very good! I missed your response when I posted my answer below. You are also correct, I should have said that the shape could be rendered geometrically in a plane or a space. But that might have given away too much.
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Luc Pattyn wrote:
you mean a hyperbole
NOW who's using hyperbole. ;P
Luc Pattyn wrote:
takes up a finite amount of space
Depends on how you define space. If you consider bounding rectangle, it takes up infinite space. If you consider the bound between the equation and the axes, I'm pretty sure that area is finite (if I felt like polishing my calculus skills, I could probably calculate exactly how much that area is). If you consider the amount of volume the curve itself displaces, it would take up no space at all. :)
I don't think you need polish, a little push will suffice, hence: The integral of
1/xisln(x) + some constant, and your function is symmetrical around the first diagonal, so the integral from 1 to infinity would cover one quarter of the total area (ignoring signs), and that quarter is infinite as it equalsln(infinity)Sorry, I can't store it without damaging it. :)Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles]
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Very good! I missed your response when I posted my answer below. You are also correct, I should have said that the shape could be rendered geometrically in a plane or a space. But that might have given away too much.
you also silently switched from edge to perimeter in your solution message; the edges get smaller and smaller after each iteration, it is the perimeter that grows to infinity. But that didn't stop me :laugh:
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you also silently switched from edge to perimeter in your solution message; the edges get smaller and smaller after each iteration, it is the perimeter that grows to infinity. But that didn't stop me :laugh:
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The solution is correct whether you use "edge" or "perimeter" as both approach infinity as the number of iterations increases. As I corrected myself below, not all edges diminish in size ;P
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The solution is correct whether you use "edge" or "perimeter" as both approach infinity as the number of iterations increases. As I corrected myself below, not all edges diminish in size ;P
Hmm. When you start with a finite triangle, and all you do is cut some edges in half, none of the edges will ever grow, let alone grow to infinity; the perimeter yes, the edges no. :)
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Hmm. When you start with a finite triangle, and all you do is cut some edges in half, none of the edges will ever grow, let alone grow to infinity; the perimeter yes, the edges no. :)
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As the number of edges approaches infinity, the sizes of those edges decreases toward 0, so they approach being a single edge. Or so, that's how I interpreted it.
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Hmm. When you start with a finite triangle, and all you do is cut some edges in half, none of the edges will ever grow, let alone grow to infinity; the perimeter yes, the edges no. :)
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Not so. You always have the outside edges of the starting triangle; those remain the same size. What you are adding are the edges created by cutting out the middle triangle, which are half the size of the triangle's outer edges.
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Q: Describe a 2 or 3 dimensional shape with an infinite edge and zero area, which takes up a finite amount of space.
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I'm surprised that no one has suggested fractals. The one I had in mind is called the Sierpinski Triangle. Observe: 1. Draw an equilateral triangle. 2. For each triangle, find the mid-point of its sides and draw lines connecting them, creating four triangles. 3. Remove the triangle in the middle, leaving three equilateral triangles connected at their vertices, each having three edges that are half of the starting triangle. 4. Go to step 2. The triangle starts with three edges of length x, so its total perimeter is 3x. After the first iteration, the shape has nine edges -- three on each of the three triangles -- each of which have a length of x/2, meaning the shape's total perimeter is 9x/2, longer than what we started with. After the first iteration, the area is 3/4 what it was before. As you continue with more iterations, the number of edges increases without bounds, and so does the shape's perimeter. As the number of iterations approaches infinity, so do the number of edges and, consequently, the length of its perimeter. Also, each iteration decreases the area geometrically: as the number of iterations approaches infinity, the area bound by the perimeter approaches zero. The shape itself never exceeds the bounds set by the starting triangle, which makes it finite. With a variant called the Sierpinski Carpet, you start with a single square, divide it into nine squares, remove the center one and repeat. Again, the number of edges and the perimeter approach infinity while the area bound by the perimeter approaches zero. There are also 3-d versions of these shapes, called sponges, which start with a tetrahedron and a cube respectively. Added: Oops, a bit of a screw up. The number of edges is doubled, not tripled; I was counting the starting edges of the triangle twice. The increase in the length of the perimeter is still 9x/2, however, as you have the three starting edges (x + x + x = 3x or 6x/2) plus the three edges of the now empty center triangle (x/2 + x/2 + x/2 = 3x/2)
modified on Thursday, May 6, 2010 8:31 PM
Here's another way to visualize it. A fractal that starts as a circle. Make a slightly smaller circle that is inside the larger circle. Line them up so the top of the smaller circle is touching the top of the larger circle. Now, the edge of that shape includes both circles (if you were to trace a pencil around the outer circle then continue on, fluidly to the inner circle). Continue making smaller circles in the smallest circle, and you end up with an infinite edge.
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mobius
---------------------------------------------------------- Lorem ipsum dolor sit amet.
That has an area.
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As the number of edges approaches infinity, the sizes of those edges decreases toward 0, so they approach being a single edge. Or so, that's how I interpreted it.
That is what I suggested here[^] but I didn't expect you to buy it. :)
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Not so. You always have the outside edges of the starting triangle; those remain the same size. What you are adding are the edges created by cutting out the middle triangle, which are half the size of the triangle's outer edges.
yes the number of edges increases; none of the new edges is larger than the three original ones. Or are we discussing something other than the Sierpinski Triangle now? You said the shape remained finite as it was limiting itself to the bounds of the original triangle. :)
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Hmm. When you start with a finite triangle, and all you do is cut some edges in half, none of the edges will ever grow, let alone grow to infinity; the perimeter yes, the edges no. :)
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Hmmm, after drawing it out, I think you are correct. While the edges get smaller, all the angles stay the same. They never really blend into a single edge. On the other hand, I still haven't been able to poke any holes in my circle fractal proposal. Feel free to do that for me. ;P
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Hmmm, after drawing it out, I think you are correct. While the edges get smaller, all the angles stay the same. They never really blend into a single edge. On the other hand, I still haven't been able to poke any holes in my circle fractal proposal. Feel free to do that for me. ;P
aspdotnetdev wrote:
I still haven't been able to poke any holes in my cirlce fractal proposal
The way I see it, it has: - no straight parts at all; - a single edge, infinitely long; - a real, non-zero, area (using the even/odd rule for inside/outside), intuitively I would say half that of the original circle. So I wouldn't call it a fractal, I wouldn't call it a solution to the problem in the OP, and frankly I wouldn't call it very pretty either. It is original though, I haven't seen it before. :)
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