| |
| |
| |
|
Page: 1 2 3 4 5 6 7 8 9
<mbot> t35t0r: I[Sqrt[1 + x^2], x]
<meh> I got an econ question
<meh> is investment a determinant of productivity?
<t35t0r> @math FindList["/etc/p***wd", "root"]
<mbot> t35t0r: No (forbidden content).
<t35t0r> @math FindList["~/.bashrc", "export"]
<mbot> t35t0r: No (forbidden content).
<t35t0r> hehe
<xerox> 'morning
<meh> where do you go for econ help?
<RyuKojiro> @math Diff[Sqrt[x^2 + 1]]
<mbot> RyuKojiro: Diff[Sqrt[1 + x^2]]
<RyuKojiro> :\
<RyuKojiro> :P
<mwc> Can somebody suggest how I'd go about proving that if the quotient group of a Group and it's Center (G / Z(G)) is cyclic, G is Abelian?
<mwc> I'm working from the angle that every coset of G is generated by (gZ(G))^n = g^n Z(G)
<mwc> and then trying to show that that implies each coset contains globally commuting elements
<mwc> I suppose an Abelian group is equal to its center, and so the quotient group would only contain itself (as identity), and this is a trivially cyclic group. Is there a way to prove a quotient group can never be cyclic if it's non-trivial?
<nerdy2> yes, show everything commutes
<nerdy2> as you said
<mwc> yeah, I can't seem to get that last bit to work out
<mwc> y g^n Z(G) = Z(G) y g^n
<mwc> if I could show that y and g^n element commuted, I'd be set
<nerdy2> what y ?
<nerdy2> hint: there is a g (your generator of G/Z(G)) such that any element of G can be written as g^k z for some k, and some z in Z(G) ...
<nerdy2> note all such things commute
<mwc> that's what I have
<mwc> Hmm, maybe I had this earlier
<mwc> does g^n Z(G) is an arbitrary coset, for the generator of the quotient group g Z(G)
<mwc> I need to figure out how to show each of those cosets consists only of commuting elements. I don't see how to do that
<thepachy> Does anyone know where I can find internet usage statistics for Philippines?
<r0b> if i have a function f(x,y) what does it mean when i need to get the total diffrential df in terms of x,y,dx,dy?
<Mulder> well
<Mulder> ***ume z = f(x,y)
<timl_> he's gone...
<Mulder> then the total differential is dz
<Mulder> oh
<Mulder> heh
<ods15> lol
<timl_> I was about to explain it too :-)
<ods15> i want to know...
<ods15> isn't total diffrential just sum of partial diffrentials?
<astrolabe_> If y = y(x) then total df/dx = partial df/dx + dy/dx * partial df/dy
<ods15> oh, only if y depends on x thugh
<r0b> if you have a function with two variables, you would have two partial derivatives right>?
<ods15> you can do partial derivative on consts, but they wouldn't be particularly interesting :P
<yyn3k> @math
<mbot> yyn3k: Null
<yyn3k> @math ?
<mbot> yyn3k: Terminated
<mbot>
<yyn3k> @math help
<mbot> yyn3k: "See http://documents.wolfram.com/v5/ for detailed Mathematica help."
<Manny> hi :)
<Manny> I'm trying to proof that n^2 <= 2^n \forall n \in N \setminus {3}
<Manny> for n=1, 2, 4 it's obvious
<Manny> now I'm doing the induction: n -> n + 1, but miserably fail :). (n+1)^2 = n^2 + 2n + 1 <= 2^(n+1)
<Manny> because of the induction precondition, I get 2^n + 2n + 2 <= 2^(n+1)
<Manny> however, now I don't know how to continue
<Manny> I tried to proof that, but it was actually wrong for n = 1
<toad-> n^2 <= 2^n => 2n^2 <= 2^(n+1), so you'd have to show that n^2 >= 2n+1
<Manny> toad-, you mean 2^n >= 2n+1
<Manny> right?
<toad-> no
<toad-> you want to show 2n^2 >= (n+1)^2
<Manny> toad-, thanks
<aj^> i have a Vector3( 20, -5, 0 ); and i want to convert it to Vector3(1,-1,0); is there a name for that ?
<Mulder> not that i know of
<aj^> i just want the direction, it seems like a typical operation, i thought there would be
<Mulder> never seen that operation before
<aj^> what do you call it when you take a number above zero, and make it +1. or if that number were -34 it would be -1.
<Mulder> dunno
<aj^> i have 2 vectors, i want the first one to change directions to point in the same direction as the 2nd vector, but not change its intensity
<OtherAlfie> The signum function gives you the sign. (I'm not sure what language you're using.)
<OtherAlfie> http://en.wikipedia.org/wiki/Signum_function
<saba> aj^, still around?
<aj^> yes.
<aj^> OtherAlfie, C++
<saba> aj^, one thing you can do is find the length of the vector A (the first one you mention), then just scale the vector pointing in the opposite direction by multiplying the unit vector by |A\
<saba> strictly speaking, suppose you have vector A and you want it to "point" in B's direction, then you want to go something like |A|/|B|*B
<saba> that should be B^
<aj^> i do not know that syntax.
<aj^> i just need to know what im doing, its 'name' so i can look it up in reference material... i dont need theory, i alreayd know what i want to do..
<saba> oh. well, suppose you have vector k. then k^ denotes the vector in k's direction with length/magnitude 1.
<aj^> i just dont know what any of this stuff is called.
<aj^> ye.s.. i know theory..
<aj^> sorry, you might not understand;
<aj^> how do i convert a vector to this vector with magnitude of 1.
<saba> divide it by its length. :D
<buggler> I LOVE SABA
<saba> HEART YOU BUGGLER
<aj^> ok.
<aj^> does the process have a name?
<saba> division.
<aj^> but specificalyl relating to vectors ?
<buggler> normalising
<buggler> oh
<buggler> my bad
<buggler> projecting
<saba> normalising sounds pretty good.
<buggler> normalising is getting u/|u|
<aj^> ok, normalizing sounds like what it is.
<buggler> projecting is getting v's component in the u direction
<genjix> is d / dx ln(4x^3 + 5x) = (12x^2 + 5) / x ?
<buggler> but oh, i see what his question was now
<saba> d/dx ln(4x^3+5x) = (12x^2 + 5)/(4x^3+5x), without simplification.
<buggler> he's doing |A| * B / |B|? Yeah, getting B / |B| is called normalizing
<buggler> ok anyway cyas everyone
<saba> toodlepip, maestro.
<aj^> saba, if its set to 0, a divide will crash
<saba> true, but then, if you have a 0 vector, all your tuples will be zero.
<aj^> thats an implementation issue, probably not appropirate for this channel.
<saba> eh, doesn't really matter.
<aj^> i have a vector3, and i want to make them all positive... does that process have a name ?
<saba> what do you mean by a positive vector?
<aj^> i mean i want all values to be set to above zero... -23,-4,2 would become 23,4,2
<aj^> im probably doing it all wrong, but i can't grasp |A|/|B|*B notation.
<aj^> i know what i want to do, but i dont know your names for these things.
<_paneb> i know how to chck to see if a set of vectors is a spanning set, but i have to do it for a set of complex vectors - is there anything different?
<none_-> is there a program that will generate algebraic equations?
<Manny> I have to chekc whether 10^x = 3 where x=p/q, and p, q \in R
<RyuKojiro> in differentiating Sqrt[x^2 + 1] I wouldn't use the chain rule, right?]
<Manny> RyuKojiro, you always use the chain rule for nested funtkions
<saba> RyuKojiro, why would you say you wouldn't?
<saba> Manny, take log base 10 of both sides? :P
<Manny> substitution is application of the chain rule
<RyuKojiro> Yeah
<Manny> saba, ouch
<RyuKojiro> But I didn't know you always used it with nested functions :p
<RyuKojiro> Thanks
<RyuKojiro> That was always iffy to me
<Manny> p, q \in Z (0, 1, -1, 2, -2, ...)
<Quiznos> hello: does the conjecture that (the sum of two odd numbers always produces an even-answer) still hold true? has it been proven false?
<Bebabo> huh??
<Bebabo> conjecture?
<Quiznos> i dont know the proper word to use in that position.
<Quiznos> is the formula true for all odd values?
<Bebabo> sure
<Quiznos> there has never been a false-positive?
<saba> z=2n+1, n integer, z is always odd.
<Quiznos> ok
<saba> so suppose you have z1 and z2 (using m).
<saba> then, z1+z2 = 2n+1 + 2m+1 = 2(n+m)+2.
<Quiznos> what about positive(z)=x+y for odd(x and y)?
<saba> this final expression is always even, irrespective of choice of n and m
<Quiznos> ok
<saba> ***uming n and m are integers, of course. :D
<Quiznos> yes
<Quiznos> anything else to add?
<Bebabo> actually you can get all rules from the Z2 field
Return to math
or
Go to some related
logs:
beginner
chatzone
replace char in c++
oracle sql convert varchar to int
|
|