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Post by Chris on Mar 7, 2006 22:54:55 GMT 1
HAHA THE LOSER IS OUTGEEKED. pi is exactly equal to twice the square of the integral from minus infinity to infinity of e to the minus x squared. Actually. It's also equal to the ratio of the circumference divided by the diameter. Surprisingly. mmm, pie. i like steak and onion pie... In fact, any n-manifold of a sphere has a ratio between the equivalents of the "circumference" and the "area" of pi/(n-1)... And Chelsea went out! WOOHOO
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Post by Simon on Mar 7, 2006 22:59:33 GMT 1
Ronaldihnio is a complete boss. silly good.
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mcv
When I Argue I See Shapes
Posts: 630
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Post by mcv on Mar 7, 2006 23:05:22 GMT 1
And Chelsea went out! WOOHOO bit of crap game though (yawn - why am i so tired?)
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Post by oasis2000 on Mar 8, 2006 1:03:52 GMT 1
WHAHAHAHA, and who put it out?? Barcelona! A Spanish team, WHAHAHAHA!!
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Post by Ben on Mar 8, 2006 1:54:34 GMT 1
HAHA THE LOSER IS OUTGEEKED. pi is exactly equal to twice the square of the integral from minus infinity to infinity of e to the minus x squared. Actually. It's also equal to the ratio of the circumference divided by the diameter. Surprisingly. mmm, pie. i like steak and onion pie... In fact, any n-manifold of a sphere has a ratio between the equivalents of the "circumference" and the "area" of pi/(n-1)... And Chelsea went out! WOOHOO mmm, topology. I've seen a graphical representation of a hypercube, but I've never seen a hypersphere. Have you? Hypercubes are much easier to define in |R^4 than a sphere, i reckon though. Indeed, the hypercube's definition comes directly from Discrete maths, where the cube graph is defined for vertices with a difference of only one co-ordinate in them, hence (1,0,0) is connected to (0,0,0),(1,1,0), and (1,0,1), which ends up forming a cube in |R^3. Sorry, rather off topic. Books anyone? Football? YAY!
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Post by Chris on Mar 8, 2006 2:24:40 GMT 1
In fact, any n-manifold of a sphere has a ratio between the equivalents of the "circumference" and the "area" of pi/(n-1)... And Chelsea went out! WOOHOO mmm, topology. I've seen a graphical representation of a hypercube, but I've never seen a hypersphere. Have you? Hypercubes are much easier to define in |R^4 than a sphere, i reckon though. Indeed, the hypercube's definition comes directly from Discrete maths, where the cube graph is defined for vertices with a difference of only one co-ordinate in them, hence (1,0,0) is connected to (0,0,0),(1,1,0), and (1,0,1), which ends up forming a cube in |R^3. Sorry, rather off topic. Books anyone? Football? YAY! Oooh, discrete maths. Plus, a nice geometric way of drawing hypercubes like so... But the whole hypesphere thing is relatively easy, if you assume that the sphere can be defined as r=sqrt(X 12 + X 22 +...+X n2) for however many dimensions, then, I dunno, it's fairly intuitive from that. Assuming that the n-area of the whole badger is (n-1)*(pi)r n/(n), which differentiates with respect to r as (n-1)*(pi)r n-1. I think.
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Post by Tom on Mar 8, 2006 2:28:30 GMT 1
im confused.
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Post by Indie Committee on Mar 8, 2006 10:06:13 GMT 1
shut up maths people.
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Post by Chris on Mar 8, 2006 15:07:32 GMT 1
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Post by The Lost Leader on Mar 8, 2006 17:23:00 GMT 1
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elaine
When I Argue I See Shapes
Mitsy the Magnificent
Posts: 605
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Post by elaine on Mar 8, 2006 19:58:04 GMT 1
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Post by Ben on Mar 8, 2006 22:08:19 GMT 1
mmm, topology. I've seen a graphical representation of a hypercube, but I've never seen a hypersphere. Have you? Hypercubes are much easier to define in |R^4 than a sphere, i reckon though. Indeed, the hypercube's definition comes directly from Discrete maths, where the cube graph is defined for vertices with a difference of only one co-ordinate in them, hence (1,0,0) is connected to (0,0,0),(1,1,0), and (1,0,1), which ends up forming a cube in |R^3. Sorry, rather off topic. Books anyone? Football? YAY! Oooh, discrete maths. Plus, a nice geometric way of drawing hypercubes like so... But the whole hypesphere thing is relatively easy, if you assume that the sphere can be defined as r=sqrt(X 12 + X 22 +...+X n2) for however many dimensions, then, I dunno, it's fairly intuitive from that. Assuming that the n-area of the whole badger is (n-1)*(pi)r n/(n), which differentiates with respect to r as (n-1)*(pi)r n. I think. yeah, i'm not saying that it's not difficult to define it, but drawing it is difficult. mapping hypercubes is easy, and indeed, drawing the version for |R^4 is relatively simple (it resembles a cube inside a rectangle with adjoint corners). I'm very tempted to make a forum for discussion of degrees; i suppose it should happen, given how we are all at university and all, and we do often do the same courses. But you messed up your differentiation; you failed to lower the power of r when you differentiated. You should have gotten (n-1)*(pi)r n-1 from (n-1)*(pi)r n/(n) not being a pedant or anything.
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Post by Indie Committee on Mar 8, 2006 23:48:34 GMT 1
i said shut it!
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elaine
When I Argue I See Shapes
Mitsy the Magnificent
Posts: 605
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Post by elaine on Mar 9, 2006 3:55:56 GMT 1
i love nat's umbrealla
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elaine
When I Argue I See Shapes
Mitsy the Magnificent
Posts: 605
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Post by elaine on Mar 9, 2006 4:01:44 GMT 1
"thereu';re like frehsers week but younhnow where to go" simon re: reading week
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emma
Backstage Pass
Posts: 64
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Post by emma on Mar 22, 2006 19:31:01 GMT 1
on the subject of comics, i thought i'd share my elation at finally finding the first collection of city of silence by warren ellis. i fucking worship that man, pretty much.
and alan moore.
hahaha, i'm way teh loooser.
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