Latest posts on math thread topichttps://rhizzone.net/forum/topic/14538/2019-04-27T04:06:42+00:00Discussion :: Laissez's Faire :: math thread (by c_man)
2019-04-27T04:06:42+00:00383932I actually mentioned that derivation to a friend of mine and he said that apparently a similar argument is made in the e&m textbook by landau and lifschitz. They're really really stellar textbooks on physics and if youre really interested you might find them interesting. Theyre a little formal but they also give a lot of intuition.
Discussion :: Laissez's Faire :: math thread (by c_man)
2019-04-26T23:25:59+00:00383929</p><style type="text/css">.custom383921{color:#0B1645 !important; background-color:#BEEBCD !important; }</style><blockquote class="custom383921"><em><a href="/forum/post/383921/">toyotathon</a> posted:</em><br/><p class="postbody_text">then i counted how many ways to make planes in each of the known (1D/2D/3D), ie how many linearly-independent 2D subspaces in each space, and saw that it matched my knowledge (0,1,3). i remembered from a prev post how the triangular numbers are related to binomials (binomials cuz i'm counting # of pairs), and made my guess for 4D/5D/6D</p></blockquote><p class="postbody_text">
<br/>I think thats probably a sensible way to understand it. What youve done, i think, is count the number of independent ways to construct an axis of rotation by constructing the number of planes normal to the axis of rotation
Discussion :: Laissez's Faire :: math thread (by toyotathon)
2019-04-26T20:20:08+00:00383921</p><style type="text/css">.custom383917{}</style><blockquote class="custom383917"><em><a href="/forum/post/383917/">c_man</a> posted:</em><br/><p class="postbody_text">This is actually a topic that has been studied in great detail. What youre describing is the representation theory of rotation groups. The result that the number of independent rotations is a triangular number can be understood as a consequence of the fact that rotations in dimension n can be represented by antisymmetric nxn matrices. The number of degrees of freedom for such matrices is exactly n(n-1)/2, which is the triangular number result.</p></blockquote><p class="postbody_text">
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<br/>okay thank you... i just learned symmetric/antisymmetric matricies in the linear algebra class so i'll reread that section and work some numbers.
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<br/>was my reasoning physical? the reasoning began, well actually it began drawing a triangle on 2 rollers, and i noticed that when it's normal force vectors were parallel, its DOF was translational, and when not, rotational. then i was like well what's a rotation, it's modulo motion... like 359.99deg-->0deg. then that, modulo isn't possible on a physical 1D line (in R1 euclidean space... like i get that a circle is 1D). it's only possible to go in one direction, and return to where one starts, in 2+ dimensions (again in flat space). and when the two vectors AREN'T parallel (= are linearly independent), they can make a plane. then i counted how many ways to make planes in each of the known (1D/2D/3D), ie how many linearly-independent 2D subspaces in each space, and saw that it matched my knowledge (0,1,3). i remembered from a prev post how the triangular numbers are related to binomials (binomials cuz i'm counting # of pairs), and made my guess for 4D/5D/6D. does that go against representation theory?
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<br/>another interesting thing i learned is that, the dimension of a space A = null(A) + rank(A). and that's probably why exact constraint works, 6 = #DOF + #constraints, say, if A is a collection of normal force vectors, then that's the # of constraints. the nullspace should be the space of motion. i think i'm close, but i don't see how it relates to my previous insight yet, about the space of the submanifold, the intersections.
Discussion :: Laissez's Faire :: math thread (by c_man)
2019-04-26T18:52:45+00:00383919Representation theory is really cool to me but the way its most commonly formulated and presented is really dull and tedious which is a real shame (how its formulated and presented varies a little depending on whether youre talking to a mathematician or a physicist but theyre both sort of blah in different ways)
Discussion :: Laissez's Faire :: math thread (by c_man)
2019-04-26T18:13:40+00:00383917</p><style type="text/css">.custom383908{color:#0B1645 !important; background-color:#BEEBCD !important; }</style><blockquote class="custom383908"><em><a href="/forum/post/383908/">toyotathon</a> posted:</em><br/><p class="postbody_text">kinda stoked on this (it's at the end)... studying exact constraint... had a 50hr train ride to read marx and think w/o any internet... remember from exact constraint that the universe has 6 degrees of freedom, 3 translation, 3 rotation.
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<br/>what always weirded me was how in 3D you get 3 rotational DOF, but in 2D, it's not 2, it's only 1. and in 1D it's zero. so: 1D/0R, 2D/1R, 3D/3R. like why the jump up by 2. in 2D you can only spin shapes around your piece of paper. and 1D there's nothing to spin into. but 3D you get 3 axes of rotation.
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<br/>well 0, 1, 3, those are the first 3 triangular numbers. and you can get triangular numbers by looking at all combinations of linearly independent vectors that span the space. so say the 3D basis is {1,0,0}, {0,1,0}, {0,0,1}. (meaning you can create any point in 3D, by adding those 3 and multiplying them by another #) so vector combinations are: {{1,0,0},{0,1,0}},{{1,0,0},{0,0,1}},{{0,1,0},{0,0,1}} = 3 total. these make planes... in 2D, say your basis is {{1,0},{0,1}}, well that's the only combo = 1 total. and in 1D, there are no combos.
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<br/>so if you believe me then 4D should be the next triangular number. 0,1,3, then 6. i got internet again and looked it up and <a href="https://en.wikipedia.org/wiki/Rotations_in_4-dimensional_Euclidean_space">IT FUCKING IS.</a> so that means 5D <a href="https://en.wikipedia.org/wiki/SO%285%29">will be 10</a>, in 6D 15 ways to spin something, etc. hella stoked that i'm not nuts thinkin bout 4D or whatever. i think these planes are actually the nullspaces of the standard rotation vectors that everybody uses to represent rotations.
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<br/>is there a linearly-independent-vector-combo equivalent for 3 vectors, like a degree of freedom which requires 3D space (like how a rotation requires 2D space)? seems like reflections work in 2D too.. like you can make a 2D reflection matrix. hell they work in 1D too by changing the sign. so not reflections...</p></blockquote><p class="postbody_text">
<br/>This is actually a topic that has been studied in great detail. What youre describing is the representation theory of rotation groups. The result that the number of independent rotations is a triangular number can be understood as a consequence of the fact that rotations in dimension n can be represented by antisymmetric nxn matrices. The number of degrees of freedom for such matrices is exactly n(n-1)/2, which is the triangular number result.
Discussion :: Laissez's Faire :: math thread (by toyotathon)
2019-04-26T07:14:16+00:00383910combining 2 unique vectors, that's unity of opposites, rotation is the result of their contradiction
Discussion :: Laissez's Faire :: math thread (by toyotathon)
2019-04-26T06:53:37+00:00383908kinda stoked on this (it's at the end)... studying exact constraint... had a 50hr train ride to read marx and think w/o any internet... remember from exact constraint that the universe has 6 degrees of freedom, 3 translation, 3 rotation.
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<br/>what always weirded me was how in 3D you get 3 rotational DOF, but in 2D, it's not 2, it's only 1. and in 1D it's zero. so: 1D/0R, 2D/1R, 3D/3R. like why the jump up by 2. in 2D you can only spin shapes around your piece of paper. and 1D there's nothing to spin into. but 3D you get 3 axes of rotation.
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<br/>well 0, 1, 3, those are the first 3 triangular numbers. and you can get triangular numbers by looking at all combinations of linearly independent vectors that span the space. so say the 3D basis is {1,0,0}, {0,1,0}, {0,0,1}. (meaning you can create any point in 3D, by adding those 3 and multiplying them by another #) so vector combinations are: {{1,0,0},{0,1,0}},{{1,0,0},{0,0,1}},{{0,1,0},{0,0,1}} = 3 total. these make planes... in 2D, say your basis is {{1,0},{0,1}}, well that's the only combo = 1 total. and in 1D, there are no combos.
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<br/>so if you believe me then 4D should be the next triangular number. 0,1,3, then 6. i got internet again and looked it up and <a href="https://en.wikipedia.org/wiki/Rotations_in_4-dimensional_Euclidean_space">IT FUCKING IS.</a> so that means 5D <a href="https://en.wikipedia.org/wiki/SO%285%29">will be 10</a>, in 6D 15 ways to spin something, etc. hella stoked that i'm not nuts thinkin bout 4D or whatever. i think these planes are actually the nullspaces of the standard rotation vectors that everybody uses to represent rotations.
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<br/>is there a linearly-independent-vector-combo equivalent for 3 vectors, like a degree of freedom which requires 3D space (like how a rotation requires 2D space)? seems like reflections work in 2D too.. like you can make a 2D reflection matrix. hell they work in 1D too by changing the sign. so not reflections...
Discussion :: Laissez's Faire :: math thread (by cars)
2019-04-06T00:44:59+00:00383104The effect of automation on workers isn't just a crisis that leads to potentially revolutionary conditions. It's probably also one of the bigger reasons people start looking into Marxism well before then based on what they've learned here and there, even if their concern doesn't start at the level of conscious thought about that specific issue.
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<br/>A lot of people, especially younger people, realize it's a dumb idea to try to suppress a lot of newer stuff that threatens people's livelihoods within the current system. They also realize that making life easier shouldn't carry the cost of making life harder. If you push hard on that contradiction and find yourself unable to accept the usual apologias, it doesn't lead to very many places besides Marxism.
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<br/>Chasing that down doesn't necessarily make someone into an effective Marxist, and it won't create a revolutionary class consciousness directly, or even contribute indirectly to it if the person's class background prevents that. But Marx & Engels, and revolutionary socialism in general, make a lot more sense on that problem than liberalism/social-democracy does, or Luddism and medieval-fetish play-acting.
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<br/>So, the problem has a lot of potential to start to change people's minds if they're not inclined toward wonky thinking about policy, since it's less an issue of one approach promising improved results over another, and more one where the usual alternatives can't really address the problem at all. It starts to make reorganizing society along different principles look like the more sensible approach, which isn't nothing.
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<br/>I don't think anyone should underestimate how much of a big deal the contradiction is for pundits and mainstream economists, either, given the vast amounts of time and money that were spent very recently trying to convince everyone that when established industry jobs fled the West, its workers would all be trained to do computers and conjure value out of thin air. It's also one of the big reasons, maybe even the biggest one, that Donald Trump's umpteenth campaign for president found mass appeal. Right around the time that even the most stubborn politicians and talking heads had to publicly accept all that stuff had fallen apart if they wanted to maintain any credibility at all, Trump started promising everyone that he'd use tariffs and threats to make water flow uphill, and he was a lot better at mustering team spirit around that promise than Pat Buchanan or whatever other cranks had been carrying the torch.
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<br/>Anyway, looking into the automation problem gives people raised on bourgeois propaganda a chance to think harder about socialism. Probably just as important, it tips people off that it won't do much if the ultimate goal of "socialism" is just higher top-bracket taxes funding social programs in Western countries, if the working class seizing the means of production is still treated as scary and profane. Remaking society to address the issue isn't as easy as reading a book and knowing it makes sense, of course.
Discussion :: Laissez's Faire :: math thread (by cars)
2019-04-05T23:27:02+00:00383102</p><style type="text/css">.custom381069{color:#0B1645 !important; background-color:#BEEBCD !important; }</style><blockquote class="custom381069"><em><a href="/forum/post/381069/">toyotathon</a> posted:</em><br/><p class="postbody_text">!!! if you can learn calc on your own you can learn <em>anything</em></p></blockquote><p class="postbody_text">
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<br/>as a complete idiot who self-taught my senior year of high school, i can't agree
Discussion :: Laissez's Faire :: math thread (by toyotathon)
2019-04-05T20:54:13+00:00383095one interesting thing about the pythagorean theorem is that it works to describe distance in any dimension(s). so hypotenuse = sqrt(x^2 + y^2) or a^2 + b^2 = c^2, and this is a visual proof <a href="https://upload.wikimedia.org/wikipedia/commons/9/9e/Pythagoras-proof-anim.svg.">https://upload.wikimedia.org/wikipedia/commons/9/9e/Pythagoras-proof-anim.svg.</a> so it describes the distance between two points, a and b. in 1D it also describes distance on the number line, by "erasing" the negative sign. so a = sqrt(a^2) = sqrt((-a)^2). it also works in 3D, like to take the string-distance between two opposite corners of a room, by measuring each length of wall and height to ceiling. this is string-distance = sqrt(L^2 + W^2 + H^2), or the <a href="https://chortle.ccsu.edu/VectorLessons/vch04/vch04_8.html">3D normal distance</a> and it works for the identity elements of addition and multiplication, 0 = sqrt(0^2), and 1 = sqrt(1^2) for whatever that's worth.
Discussion :: Laissez's Faire :: math thread (by toyotathon)
2019-03-08T23:18:08+00:00381493it's good to have a good reason to go to class. like you're learning it not for an abstract obligation to be in school, but because you need to solve a specific problem or job. it's also good if that specific problem/job intersects w/ your ideology some way. like i took machining classes cuz i needed to make a part on the lathe then i left after i was done figuring out the lathe* & how to make my part. there were some gunsmith guys in there basically doing that too like rifling barrels or whatever
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<br/>*i was not
Discussion :: Laissez's Faire :: math thread (by drwhat)
2019-03-07T22:53:35+00:00381435As someone who fucks up various things regularly due to weird anxieties about class attendance, here is what I would need to hear: get over your shit and go. It doesn't matter how awkward or stupid you think you are or what bad encounter you're envisioning, it will only last a few minutes at most and learning something you want to learn is way more important to your entire life than one little bullshit moment where you'll feel weird and bad for a bit. Then you can just get on to the learning part. Please do it.
Discussion :: Laissez's Faire :: math thread (by Belphegor)
2019-03-04T15:50:52+00:00381268Y'all are so cool, I'm fascinated by the work and study everyone is doing.
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<br/>I really want to do interesting technical work but I'm having trouble breaking in to the field. I got a machinists apprenticeship last year but I lived downtown and the factory was way out in suburbia. Since I don't drive, I had a really hard time making it work and I couldn't keep it up.
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<br/>Now I'm in a technicians program in college. My challenge now is the paranoia and anxiety of being in this massive crowded school. Classes are trivial so far but I keep skipping them to avoid the bus/the halls.
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<br/>I don't want to stink up the math thread with my gripes, I just needed to share these woes with people who will understand .. thank u
Discussion :: Laissez's Faire :: math thread (by tears)
2019-03-03T22:53:17+00:00381239</p><style type="text/css">.custom381236{}</style><blockquote class="custom381236"><em><a href="/forum/post/381236/">nearlyoctober</a> posted:</em><br/><p class="postbody_text">read john smith's <em>imperialism</em></p></blockquote><p class="postbody_text">
<br/>you might consider rhizzone favorite "divided world divided class" by zak cope if you have not already
Discussion :: Laissez's Faire :: math thread (by nearlyoctober)
2019-03-03T22:04:04+00:00381236yea, i agree with everything here. i'm shouldn't be so delusional to expect that i'll ever have to worry about these questions personally... besides, what you say:
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<br/></p><blockquote><p class="postbody_text">so the material pre-conditions of socialism in textile-making countries probably aren't CNC garment machines, it's probably the destruction of the US navy, or at least that'd be a good start.</p></blockquote><p class="postbody_text">
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<br/>is tragically true. the technologies of socialism are clearly overripe, and they're actually rotting in under-utilization and misuse. so i have to apologize because my questions were selfish. but that was a fascinating article and your comments were very helpful so not all is lost. you've also reminded me i should read john smith's <em>imperialism</em>. i never finished it because i got so lost in the methodology but i bet i could tackle it now