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Comments:
<0> take the set of all subatomic particles in this universe, now add one, they are not 1-1
<1> that's because you're not using math to understand the physical world
<2> It doesn't need to be, unless you want to show that a part of math is useful.
<1> once you start doing that
<1> you better hope it's true
<3> yeah, but if math says 1==2 is false, and nature says otherwise, does not mean math is WRONG
<3> which is my argument
<2> Math doesn't say that.
<1> i never said it did.
<0> math doesnt have to be proven, it's just another language, that's like saying i need to prove that english is right, but it's just a means to communicate ideas
<2> The integer axioms might say that.
<1> math is not a language.
<0> yes it is.
<2> If the integer axioms say that and nature says otherwise for some phenomenon, that phenomenon doesn't use integers.
<1> and i never said that natural truths imply mathematical truths
<1> but if you want to use mathematics to understand nature, then some of those truths had better hold in nature, otherwise you're just tilting at windmills
<2> There's no real right or wrong about math in that sense. Either a part of math is applicable to a part of nature or it's not.
<4> I am doing some problems involving the properties of radicals and the instructions are to simply radicals by rationalizing the deonomnators.
<0> simplify?
<4> Denominator: ^of X - ^ 2 Would this invole multiplying ^ ofx - ^ of 2?
<2> What is ^ of X ?
<4> square root of x.
<2> Sqrt[x]
<3> sqrt(x)
<2> Multiply it by itself with the sign in the middle changed.
<4> Is the sign in the middle always the opposit of what was previous If it is sqrt(y) + 3 would you multiply it by sqrt(y) - 3??
<4> The book wasn't clear on when to use a + or a minus.
<5> Defender: Yes. You want to remove any square roots.
<5> [sqrt(y) + 3][sqrt(y) - 3] = y - 9
<5> [sqrt(y) + 3][sqrt(y) + 3] = y +6sqrt(y) + 9
<4> I see. so the sign in the middle is always the opposite of the equation you are working with.
<0> Defender, think of it as a quadratic
<0> what would you do to make the middle term cancel out
<4> ok
<1> Defender: yes. the underlying idea is difference of squares: (a-b)(a+b) = a^2 - b^2
<4> How do I simplify [16 1/3*x4^3] + [5 * 2^1/3*x^1/3] ???
<1> i'd start by factoring out an x^(1/3)
<6> whats the space doing betweeb 16 and 1
<6> and you could use some parentheses
<5> Defender: I think you need more parentheses. I can't follow.
<0> exponentiate it, no?
<6> not hard to follow but the chance is we do what you aint trying to do
<5> Which is why I can't follow it.
<5> I don't know what he means.
<4> I have a camera. Could you see it on that?
<6> i could, just use proper operator priority, but i aint gonna waste time when it might not be what he wants
<4> I am trying to simplify [(16^1/3)*(x^4/3) + (5*2^1/3*x^1/3)]
<5> Defender: is that (x^4)/3 or x^(4/3)?
<4> NO
<5> I read x^4/3 as the former.
<5> No.
<7> anyone here have a name of a really good book of math for High School, but a really good book (or books) ?
<4> It is sixteen to the 1/3rd power
<4> times x to the 4/3rd power
<0> Adriel_BR, check out the schaums line of books
<0> lots of details, straight to the point
<7> frostburn, thak you
<0> cheap too =)
<7> good!
<5> So what you mean is 16^(1/3) * x^(4/3) + 5*2^(1/3) * x^(1/3), correct?
<0> they're carried in any store like borders, barnes and noble etc
<4> YES
<5> So write it that way next time. We can't just know what you mean.
<0> wheres my ti93...
<0> 2*
<4> Sorry I haven't taken a course yet on Math pigmy sink.
<4> So how do i factor this ****er out?
<5> So factor out the x^(1/3). x^(1/3)[2*2^(1/3)*x + 5*2^(1/3)].
<4> So far I am getting 16^(1/3) + 5*2^(1/3)x^(1/3)
<5> Now factor out the 2^(1/3). (2x)^(1/3)(2x+5)
<5> You just dropped the x^(4/3) factor.
<5> "Math pigmy sink"?
<4> well may be I did something wrong. I really don't follow you.
<4> There is no 2x^(1/3). It is 2^(1/3)x^(1/3)
<5> I didn't say 2x^(1/3), I said (2x)^(1/3).
<4> Sorry I don't have a 2x. I only have a 2^(1/3)
<4> You are combining variables and constants
<4> Not like terms
<5> 2^(1/3) * x^(1/3) = (2x)^(1/3)
<5> % Simplify[ 2^(1/3) * x^(1/3) == (2x)^(1/3) ]
<8> |Steve|: True
<4> We haven't covered that theorem yet. All we have covered is a^m + a^n=a^m+n
<4> The original problem was combine the radical expressions of cubert of (-16x^4) + 5xcubert of 2x
<5> Defender: That's not correct.
<5> cubert?
<4> means cube root of -->
<5> a^m * a^n = a^(m+n). You really need parentheses.
<4> It would be a lot easier if you could look at what is on paper or in my book for problems. I don't know Latex and mathbin yet, but plan on learning it.
<0> log, log, it's big it's heavy it's wood...
<5> You can type it in text as long as you do it in an unambiguous way.
<4> Obviously I am not doing a very good job because no one but me knows what I mean.
<5> Well, how are we supposed to know that a^n+m means a^(n+m) rather than (a^n)+m?
<9> anybody done previous research on the interior product in a gr***man algebra?
<9> i'm looking for techniques i can use to show that the intersection of ker i_x and ker i_y is non-trivial for particular x and y
<4> I have a camera. It would be a lot easier if you would just allow me to point it at the book. :)
<5> Just use parentheses!
<10> slava: would have to totally switch gears and start thinking in those terms again, what are you trying to do? =)
<10> slava: has been a long time since I did anything there =)
<5> Learning the order of operation would be enough to show you where you need to use it.
<9> edwardk: actually, i'm trying to show that the cohomology of a nilpotent lie algebra admits at least one non-trivial cohomology operation
<10> slava: what particular x and y?
<9> suppose x and y belong to /\^2 V.
<10> slava: little outside of my area of expertise. mostly used it for computational geometry stuff
<4> Problem is combine the radical expressions: [cube root of (-16x^4)] +[5x*cuberoot of 2x]
<5> /\^2V?
<9> second exterior power
<9> of your vector space
<5> Ah.
<9> /\ is a Lambda :)
<5> Well, what you're doing is beyond me. I never learned cohomology. Nor lie algebras for that matter.
<5> I hope we get into Lie groups and such next quarter though.
<9> well i've reduced my problem to learning more about the interior product
<9> right now i'm looking at two-step lie algebras
<5> Defender: I think I already solved it all above.
<9> and the laplacian reduces the task of finding cohomology operations to looking at the kernel of i_x and multiplication by x for various x
<11> what is i_x ?
<10> my brain always shut down when people started talking cohomology, but then I never took any cl***es in homological algebra, etc.
<9> the adjoint under the inner product of left multiplication by x in the gr***man algebra
<9> i_x(x) = 1
<9> i_x(y) = 0 if x and y are l.i.
<9> i_x(a*b)=i_x(a)b +/- a*i_x(b) for higher degree elements, its a derivation
<9> kernels of derivations seem pretty tricky to get a concrete feel for
<11> all i know about derivvations is for groups, and there they corresond to half of a homomorphism on a semidirect product
<9> in a ring a derivation is a map satisfying d(uv)=(du)v + u(dv)
<11> yes, it's the same as for groups
<9> so in the ring of C^infty functions you have the derivation which corresponds to ordinary differentiation
<9> but in my case the ring is a finite dimensional algebra over the rational numbers
<11> derivations are related to extensions, so that would seem to be a way to understand the kernel, via extensions
<9> and the derivation satisfies d^2=0
<12> what's the character before phi in the greek alphabet?
<11> epsilon?
<13> ypsilon or upsilon, not epsilon
<12> upsilon according to wikipedia
<1> yes. epsilon is way earlier.
<12> my latex cheat sheet isn't in (greek) alphabetical order ;/
<14> you math geeks never got to sleep ;)
<14> go*
<12> ;)
<5> ldexter: Well, we're all over the world.
<14> i knew that was coming
<14> i don't want to hear any excuses
<12> some of us just prooved (I hope) the google ***ignment (something that appears on an exam google gives, according to my prof.)
<14> congratulations :-)
<12> thanks ;)
<10> myrkraverk: there was an old google thing that wolfram went through and wrote up how to solve everything on it in mathematica a while back
<10> well, more likely given what i hear, wolframs' lackeys did and he wrote it up as if he did it himself... ;)
<12> edwardk: well, I was quite explicitly told not to google for answers ;)
<10> myr: heh
<12> edwardk: so I'm allowed to be proud of myself, until my prof. tells me it's wrong ;)
<10> hehehe
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