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Comments:
<0> A*A doesn't make sense AxA does, but that's not important here. P(A) is the powerset of A.
<0> If |A|=8, then the number of subsets containing exactly 5 elements is 8 choose 5 = 56.
<1> % Expand[(2k+1)^2]
<2> Mulder: 1 + 4*k + 4*k^2
<0> norton: You're wrong.
<3> norton: imaginary
<4> % Solve[2^x == -2, x]
<2> Catfive: {{x -> (I*Pi + Log[2])/Log[2]}}
<0> 4^x != (2^x)^2 = 2^(2x).
<0> % {4^3,(2^3)^2,2^(2*3)}
<2> |Steve|: {64, 64, 64}
<0> Hmm, bad example, apparently.
<0> % {4^5,(2^5)^2,2^(2*5)}
<2> |Steve|: {1024, 1024, 1024}
<0> And I'm just wrong or something.
<4> norton is not wrong.
<4> % 4^x - 2^(x+1) - 8 == 0 /. x -> 2
<2> Catfive: True
<5> \o/
<0> heh
<4> % 4^x - 2^(x+1) - 8 == 0 /. x -> (I Pi + Log[2])/Log[2] //FullSimplify
<5> :)
<2> Catfive: True
<5> but I don't understand really why I get 2^x = -2
<1> % Factor[2k + 2k^2 + 1/4]
<2> Mulder: (1 + 8*k + 8*k^2)/4
<4> norton - since -2 is a perfectly good solution to your original quadratic equation.
<5> Catfive: you mean t^2-2t-8 = 0?
<4> or the one before that, same thing
<5> really? If I x=-2 in (2^x)^2-2*2^x-8 I don't get 0.
<0> 2^x = -2.
<5> oh, sorry, my mistake
<0> % Solve[2^x==-2,x]
<2> |Steve|: {{x -> (I*Pi + Log[2])/Log[2]}}
<1> % Solve[2^x (1 + 2.2^x) == (y+1)(y-1), x]
<2> Mulder: Solve[2^x*(1 + 2.2^x) == (-1 + y)*(1 + y), x]
<1> % Solve[2^x (1 + 2.2^x) == (y+1)(y-1), {x,y}]
<2> Mulder: {{y -> -1.*Sqrt[1. + 2.^x + 2.^(2.137503523749935*x)]}, {y -> Sqrt[1. + 2.^x + 2.^(2.137503523749935*x)]}}
<5> ok, I understand it better now :) thanks for the help!
<6> when a series is n, (n-1)+4, ... i.e. 0,4,8,12,16
<6> is this said to have arithmetic progression?
<7> is ((A union B) intersection C) = (A union (B intersection C)) ?
<0> No.
<0> What happens if A is not the empty set and C is thet empty set?
<0> (Doesn't matter what B is.)
<8> it's ((A intersection C) union (B intersection C))
<7> if C is empty then the first expression will be empty
<0> But will the second?
<7> nope
<0> So can they be equal?
<7> not if C is empty
<0> So that answers the question.
<7> but it some cases they're equal
<7> in
<0> So?
<8> if A is not a subset of C, the equation will be false.
<1> % Solve[1 + 2^x + 2^(2x+1) == y^2, x,y]
<2> Mulder: Solve[1 + 2^x + 2^(1 + 2*x) == y^2, x, y]
<1> % Solve[1 + 2^x + 2^(2x+1) == y^2, {x,y}]
<2> Mulder: {{y -> -Sqrt[1 + 2^x + 2^(1 + 2*x)]}, {y -> Sqrt[1 + 2^x + 2^(1 + 2*x)]}}
<8> and if A is a subset of C it will be true, because then A intersection C = A.
<0> Listen to int-e on this one.
<1> % Solve[1 + 2^x + 2^(2x+1) == y^2, x]
<2> Mulder: {{x -> Log[(-1 - Sqrt[-7 + 8*y^2])/4]/Log[2]}, {x -> Log[(-1 + Sqrt[-7 + 8*y^2])/4]/Log[2]}}
<7> are you guys good at drawing Venn-diagrams, I'm not sure how to depict X = ((A union B) intersect (complement of C)) where the universal set is U
<0> matekuten: Draw three intersecting circles and then shade the appropriate region.
<0> In this case, the parts of A and B that don't intersect C.
<7> so I just draw a circle C outside A and B somewhere?
<0> No.
<0> Make them all intersect.
<7> ok
<7> oh right
<0> http://raoulhl.dyndns.org/~steve/venn.png
<7> cool
<7> calculate the set M(1) = {0, -2/3, 4/5, -6/7. 8/9, -10/11,...) with the principle of inclusion
<0> What do you mean "calculate the set"?
<7> well name the set
<7> not calculate
<7> state, give, mention
<7> indicate, show, note
<7> specify
<7> it's one of those I'm sure
<9> i'd name it harold
<10> hi
<10> what is discrete mathematics
<0> How about M(1) = { (-1)^k * 2k/(2k+1) | k \in N } where I'm including 0 in N.
<10> hi
<4> pawan1234, are you stuck in a time warp?
<10> no
<10> finding hard
<10> discreet
<10> discrete
<4> must we go through this same sequence of events over and over, like some particularly banal episode of the Twilight Zone that didn't make public screening?
<0> hi
<0> (Sorry, couldn't resist.)
<7> aren't all the episodes of twilight zone pretty much the same?
<7> This guy wakes up and evryones changed except him
<10> what
<7> |Steve| that doesnt make much sense to me
<11> what exactly is a supertask?
<0> I didn't use the principle of inclusion.
<10> dijikstra algorithm
<0> As I'm not sure what that is, I used set comprehension.
<7> well it says rules of inclusion
<0> pawan1234: Try using some verbs.
<11> wikipedia says it's something done with infinitely many steps in a finite time. but "time" isn't defined. looking at the result of infinitely many steps, but still regarding the process as having taken infinite time, would still be considered a supertask, right?
<11> mathworld is down and i don't find anything but wikipedia mirrors when i google for supertask
<12> Why are there so many more people in #math than in #physics
<0> Failure02: Try #philosophy
<0> No one talks in #physics. I stopped going.
<11> |Steve|: why?
<12> |Steve|: right, but that means you contributed to its lack of people
<0> Failure02: Well, because wikipedia says, "In philosophy, a supertaks is..."
<0> JohnFlux: Indeed.
<0> Plus, I finished my physics minor and didn't need any help (not that they ever gave any).
<0> You know, that might have been #physics on efnet though.
<11> |Steve|: so? hilbert's hotel is a real math problem, but still a supertask.
<0> In a way, I suppose.
<7> |Steve|: what's N in what you wrote earlier?
<0> That page has an odd example of Hilbert's paradox. It's usually move to the next room.
<0> matekuten: The natural numbers.
<7> oh right
<11> which page?
<7> I can
<11> the hilbert's hotel page?
<6> heh, Absolute Steve
<11> that wikipedia article is pretty crappy
<0> Failure02: I'm not saying not to ask here. I'm just saying that you might find more people who know about it is all.
<7> I can't write: k is a positive integer?
<0> The supertask page.
<0> matekuten: You need zero.
<0> You can say nonnegative integer.
<7> I can't write: k is a positive integer >= 0?
<0> No.
<0> 0 is not a positive intenger.
<9> lies
<12> say "non-negative"
<0> No need for the hyphen.
<11> well it sounded a bit like an insult, like "that's not math. go away". if it wasn't, then nevermind. :)
<7> well if I just say k is an integer >= 0
<0> Oh, I'm sorry. I didn't mean to imply that at all Failure02.
<11> alright.
<0> matekuten: You shouldn't write >= in the middle of a sentence unless you have something other than prose on both sides.
<0> er
<7> really?
<0> Don't try to work out the negatives in that sentence.
<13> what is wrong with that?
<0> Say, "k is an integer at least 0" or "k>=0."
<0> Saying "k is an integer >= 0" is sloppy.
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