EPT2) 4*3
3) 12-9
4) 14/7
5) How many two dimensional surfaces does a seven dimensional (hyper) cube have? Justify your answer.
Math Homework Help
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micronesia
United States24745 Posts
2) 4*3 3) 12-9 4) 14/7 5) How many two dimensional surfaces does a seven dimensional (hyper) cube have? Justify your answer. | ||
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EmeraldSparks
United States1451 Posts
2. 42 3. 42 4. 42 5. 42 | ||
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MeriaDoKk
Chile1726 Posts
2.-15 3.-7 4.-67 5.-2 | ||
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Boblion
France8043 Posts
2) 12 3) 3 4) 2 5) + Show Spoiler + ![[image loading]](http://pix.nofrag.com/b/e/8/9f5e3204ed557d25b6473bb313417.jpg) ![[image loading]](http://pix.nofrag.com/7/a/b/777b192f5af6b4e588bba2c12beab.gif) | ||
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Lemonwalrus
United States5465 Posts
2. 12 3. 3 3. 2 5. 672. (E sub m,n = 2 to the n-m * (n! / m!(n-m)!) where n equals cube dimensions and m equals surface dimensions.) | ||
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Boblion
France8043 Posts
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micronesia
United States24745 Posts
On April 19 2008 12:28 Lemonwalrus wrote: 5. 672. (E sub m,n = 2 to the n-m * (n! / m!(n-m)!) where n equals cube dimensions and m equals surface dimensions.) Is that what you get when the take the matrix: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 23 7 1 and square it? | ||
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Lemonwalrus
United States5465 Posts
 Edit: And I don't even remember how to do the simplest of calculations using matrices. | ||
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Boblion
France8043 Posts
[spoiler][spoiler][spoiler][spoiler]+ Show Spoiler + + Show Spoiler + you are mocking him ? | ||
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micronesia
United States24745 Posts
On April 19 2008 12:33 Lemonwalrus wrote: I just found a formula to solve the problem, the actual math/reasoning involved is way above my head, sorry. Tomorrow when I have time I'll run through the calculations for my method to check if you are correct (unless someone else comes here and gives a convincing response) | ||
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Lemonwalrus
United States5465 Posts
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azndsh
United States4447 Posts
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micronesia
United States24745 Posts
On April 19 2008 12:36 Lemonwalrus wrote: What level math is this for? It's just satire but I didn't want to make a COMPLETELY insubstantial post so I added a random math procedure I vaguely understand (with no other exigency for posting it). azndsh I think you need to explain that.. | ||
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Lemonwalrus
United States5465 Posts
On April 19 2008 12:37 micronesia wrote: azndsh I think you need to explain that.. Second. | ||
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azndsh
United States4447 Posts
Thus, one of these squares would be the following four points (0, 0, 0, 0, 0, 0 ,0) (1, 0, 0, 0, 0, 0, 0) (0, 1, 0, 0, 0, 0, 0) (1, 1, 0, 0, 0, 0, 0) Notice that only the first 2 coordinates are different. Its easy to see that is true for any unit square. I could describe this square as (x, x, 0, 0, 0, 0, 0) Thus 5 coordinates are fixed, while the remaining 2 describe the square. There are (7 choose 5) * 2^5 ways to fix 5 coordinates The other formula corresponds to the general case. | ||
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micronesia
United States24745 Posts
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Lemonwalrus
United States5465 Posts
Edit: (7 choose 5) * 2^5 = 672, right? | ||
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azndsh
United States4447 Posts
you can extend the argument to arbitrary (m,n) | ||
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micronesia
United States24745 Posts
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Lemonwalrus
United States5465 Posts
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micronesia
United States24745 Posts
On April 19 2008 12:36 azndsh wrote: if you fix 5 coordinates, the remaining 2 coordinates determine a unit square... thus 632 ^there | ||
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azndsh
United States4447 Posts
definitely 672 | ||
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micronesia
United States24745 Posts
On April 19 2008 16:10 azndsh wrote: typo, sorry hehe definitely 672 All right, cool. I know how to calculate the answer to it (I showed the work above, sort of) but I have absolutely no understanding of it. | ||
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sigma_x
Australia285 Posts
Sum( C(n,k)C(k,m) ) = 2^(n-m) C(n,m) for an "n" dimensional cube within which we select "m" dimensional cubes. So, for example, a 3-dimensional cube has 6 2-dimensional cubes (faces). We shall prove the identity by double-counting. The right hand side is readily solvable by graph theory. Each n dimensional cube has 2^n vertices. Now, there are C(n,m) ways of selecting an m-th dimensional cube. Further, each selected m-th dimensional cube has 2^m vertices, so that we have 2^n * C(n,m)/2^m and we are done. For the second method, place the n-th dimensional cube on a Cartesian plane. Let the cube have side length 2. Then each vertex can be specified by (±1,±1,±1,...,±1,±1). Now, for any given vertex defined by a set of coordinates, we can change m of them (from +1 to -1 and vice versa) and the vertex will still lie on the m-th dimensional cube. Further, these two points define a unique m-cube (since this is the maximal number of coordinates that can be changed for both coordinates to lie on the same m-cube). Now, there are C(n,m) ways of selecting m coordinates (and assign them the value +1). For each of these, m of them can be changed to -1 C(m,m) ways. Together, these two points define a unique m-cube. Thus, there are C(n,m)C(m,m) ways of doing this. Now, do the same by assigning m+1 coordinates +1, C(n,m+1) ways. For each of these, m of them can be changed to -1, C(m+1,m) ways. Thus, there are C(n,m+1)C(m+1,m) ways of doing this. We repeat this process until we are assigning n coordinates +1, of which we change m of them to -1. Since this process is mutually disjoint, and covers all m-cubes, the total number of m-dimensional cubes in an n-dimensional cube is Sum( C(n,k)C(k,m) ). This completes the proof and explains why your process works. Edit: lol, i didn't actually answer the question. Let n=7, m=2, and we have C(7,2)C(2,2)+C(7,3)C(3,2)+C(7,4)C(4,2)+C(7,5)C(5,2)+C(7,6)C(6,2)+C(7,7)C(7,2)=21x1 + 35x3 + 35x6 + 21x10 + 7x15 + 1x21 = 2^5 C(7,2) = 672 | ||
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micronesia
United States24745 Posts
On April 20 2008 16:49 sigma_x wrote: What your method exhibits is the following: Sum( C(n,k)C(k,m) ) = 2^(n-m) C(n,m) for an "n" dimensional cube within which we select "m" dimensional cubes. So, for example, a 3-dimensional cube has 6 2-dimensional cubes (faces). We shall prove the identity by double-counting. The right hand side is readily solvable by graph theory. Each n dimensional cube has 2^n vertices. Now, there are C(n,m) ways of selecting an m-th dimensional cube. Further, each selected m-th dimensional cube has 2^m vertices, so that we have 2^n * C(n,m)/2^m and we are done. For the second method, place the n-th dimensional cube on a Cartesian plane. Let the cube have side length 2. Then each vertex can be specified by (±1,±1,±1,...,±1,±1). Now, for any given vertex defined by a set of coordinates, we can change m of them (from +1 to -1 and vice versa) and the vertex will still lie on the m-th dimensional cube. Further, these two points define a unique m-cube (since this is the maximal number of coordinates that can be changed for both coordinates to lie on the same m-cube). Now, there are C(n,m) ways of selecting m coordinates (and assign them the value +1). For each of these, m of them can be changed to -1 C(m,m) ways. Together, these two points define a unique m-cube. Thus, there are C(n,m)C(m,m) ways of doing this. Now, do the same by assigning m+1 coordinates +1, C(n,m+1) ways. For each of these, m of them can be changed to -1, C(m+1,m) ways. Thus, there are C(n,m+1)C(m+1,m) ways of doing this. We repeat this process until we are assigning n coordinates +1, of which we change m of them to -1. Since this process is mutually disjoint, and covers all m-cubes, the total number of m-dimensional cubes in an n-dimensional cube is Sum( C(n,k)C(k,m) ). This completes the proof and explains why your process works. Edit: lol, i didn't actually answer the question. Let n=7, m=2, and we have C(7,2)C(2,2)+C(7,3)C(3,2)+C(7,4)C(4,2)+C(7,5)C(5,2)+C(7,6)C(6,2)+C(7,7)C(7,2)=21x1 + 35x3 + 35x6 + 21x10 + 7x15 + 1x21 = 2^5 C(7,2) = 672 Every time someone attempts to explain this to people who haven't learned about it already, it seems to result in them going 'HUH?' I guess only those who study the proper topics in college will get what the hell this means. | ||
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