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Comments:
<0> thanks
<1> % DSolve[{y'[x] == x^2 - 1/y[x], y[1] == -2}, y[x], x]
<2> doojin: "Time limit exceeded for computation."
<1> oops
<1> % DSolve[y'[x] == x^2 - 1/y[x], y[x], x]
<2> doojin: "Time limit exceeded for computation."
<1> what's wrong?
<1> % D[y[x] == x^2 - 1/y[x], x]
<2> doojin: Derivative[1][y][x] == 2*x + Derivative[1][y][x]/y[x]^2
<1> jesus
<3> hi
<3> does anyone know if SPSS can do clustered stacked charts?
<4> ****, vector space nor linear algebra on wikip has links to Rank! can you believe that?
<5> Huh?
<5> http://wikipedia.org/wiki/Rank_(linear_algebra)
<4> Catfive: you didn't read me carefully i don't think
<5> I don't understand you. Rank is not a property of a vector space, and it's specific enough that it's hardly surprising the generic 'linear algebra' article wouldn't mention it.
<4> Catfive: so, what i said in the first time is perfect?
<1> how do I write ln(x)?
<1> in mathematica
<6> % D[Log[x], x]
<2> Mulder_: x^(-1)
<1> Log[x]?
<1> hmm
<1> % PlotVectorField[{1, -Log[x]}, {x, -2, 2}, {y[x], -2, 2}]
<2> doojin: No (forbidden content).
<1> it doesn't work in mathematica
<6> Log[x] is ln(x) in mathematica
<1> I know
<1> PlotVectorField[{1, -Log[x]}, {x, -2, 2}, {y[x], -2, 2}] makes errors
<1> oh man
<1> I need to set the range of x bigger than 0
<7> TBA (topic)?
<5> kmh, presumably whenever he feels like it.
<7> i mean TBA=?
<5> 'to be announced'
<7> ah
<7> ty
<5> @arrr
<2> Avast!
<6> % Solve[ c == a*sin[x] + b*cos[x], x]
<2> Mulder_: Solve[c == b*cos[x] + a*sin[x], x]
<6> % Solve[ c == a*Sin[x] + b*Cos[x], x]
<2> Mulder_: {{x -> -ArcCos[(c - (a^2*c)/(a^2 + b^2) - (a*Sqrt[a^2*b^2 + b^4 - b^2*c^2])/(a^2 + b^2))/b]}, {x -> ArcCos[(c - (a^2*c)/(a^2 + b^2) - (a*Sqrt[a^2*b^2 + b^4 - b^2*c^2])/(a^2 + b^2))/b]}, {x -
<2> > -ArcCos[(c - (a^2*c)/(a^2 + b^2) + (a*Sqrt[a^2*b^2 + b^4 - b^2*c^2])/(a^2 + b^2))/b]}, {x -> ArcCos[(c - (a^2*c)/(a^2 + b^2) + (a*Sqrt[a^2*b^2 + b^4 - b^2*c^2])/(a^2 + b^2))/b]}}
<8> quite simple then
<6> lol
<9> % Solve[c == a Sin[x] + b Cos[x], x] // FullSimplify
<2> Olathe: {{x -> -ArcSec[(b*(a^2 + b^2))/(b^2*c - a*Sqrt[b^2*(a^2 + b^2 - c^2)])]}, {x -> ArcSec[(b*(a^2 + b^2))/(b^2*c - a*Sqrt[b^2*(a^2 + b^2 - c^2)])]}, {x -> -ArcSec[(b*(a^2 + b^2))/(b^2*c + a*
<2> Sqrt[b^2*(a^2 + b^2 - c^2)])]}, {x -> ArcSec[(b*(a^2 + b^2))/(b^2*c + a*Sqrt[b^2*(a^2 + b^2 - c^2)])]}}
<6> so trivial *cough*
<9> Looks like +/- ArcSec[(a^2 + b^2)/(b c +/- a Sqrt[a^2 + b^2 - c^2])]
<10> hello....
<10> someone knows the formula to get the area of a polar function?
<10> imean, a function in polar intervals
<10> ?
<7> hmmm
<7> what exactly are you trying to do ?
<10> i'm studying integrals
<10> i need to get the area of a integral in polar intervals
<10> but i don't remember the formula
<7> i'm not sure i get that
<10> Integral[0,Pi] f(x)
<10> but isn't just f(x), must be f(x)^2 or some
<7> F[Pi]-F[0]
<7> with F'(x)=f(x)
<10> just that?
<7> just that
<10> so, i must do an Integral in 0,pi of F'(x)dx ?
<7> ?
<10> http://www.mathwords.com/a/area_polar.htm
<10> there is :D
<7> int(f(x),x=0..Pi)=F(Pi)-F(0)
<7> oh
<7> that is the sector of 2d curve
<10> 1/2int(r^2, x=0,x=Pi)
<7> for that you can use the leibniz formula
<10> but if you change a=0 and b=pi it works
<10> i need to do it by integrals
<7> i don't know what you mean by "need to do it by integrals" ?
<10> mmm... i need to do the formula 1/2int(r^2, x=0,x=Pi)
<10> it's ok, i did find it
<10> thanks :)
<7> that formula won't work always
<10> when don'pt works?
<7> nvm just use that, i can't find an online reference to the sector formula
<11> I have a week from now to become one with generating functions!
<7> Copter : there's a free book/pdf by herbert wilf
<7> http://www.cis.upenn.edu/%7Ewilf/
<11> Cool , anything nice that I can look at and find ideas and improve my thinking :)
<7> first link on the page
<11> got it thx
<12> hi
<12> what is 1/9^-2?
<12> numerically
<13> 81
<12> what is 1/3*9^-2?
<12> one example I have says 27
<13> use more parentheses
<13> is 9^-2 in the denominator?
<12> yes and there are no parentheses
<13> if it's in the denominator you need to write that as 1/(3*9^-2) then
<13> and it's 27, yes
<12> what is the order of operation to get that?
<13> well 9^-2 = 1/9^2 so 1/(3*9^-2) = 9^2/3 = 27
<12> ifthanks
<12> if I first take the 9^-2 in the denominator I get 1/81
<12> if I multiply that by 3 I get 1/3/81
<12> I see
<14> what does 1/3/81 mean?
<14> you need parentheses
<14> either say 1/(3/81)
<14> or (1/3)/81
<14> they are different
<14> toko123: ^
<7> dr Flux
<7> how's aeon ?
<12> thanks for the help
<7> omg vogons
<7> get your towels and jump
<5> Here is a simple but fun problem; what is your favourite solution? Suppose we're working in a normed vector space. Show that B_r(v) + B_s(w) = B_{r+s}(v + w), where B_r(v) is the open ball of radius r about v.
<15> translate both sides by -v-w, then apply obvious properties of the norm.
<16> hello
<16> can someone just clarify exactly what a kernel is in linear algebra?
<5> longodj, if f: G -> H is any group homomorphism, the kernel of f is the set {a in G : f(a) = e}, where e is the group identity.
<5> longodj, in particular, if V and W are vector spaces and f: V -> W is a linear transformation, ker(f) = {v in V : f(v) = 0} is the set of vectors in V that f sends to 0.
<16> that's what I wanted
<16> thanks a bunch
<16> :)
<16> I kept seeing a bunch of math jargon....i should understand it but I'm a bit tired....so I was looking for that goes to zero business :)
<16> thanks
<7> Catfive : hmmm it looks odd at a first glance
<5> kmh - sorry?
<7> your fun problem
<5> how do you know it's odd and not even? =)
<7> if i visualize it (wrongly i ***ume) it makes no sense
<7> it looks like the union of 2 arbitrary circles is a circle (which is false)
<5> it isn't a union, though, it's a sum.
<7> oh
<7> i guess that was the first misinterpretation
<7> that explains the problem with that visualization
<7> lol
<7> whatever but inverse ?
<17> well whyever not? :)
<7> just asking :)
<17> my new gentoo box's hostname is euclid
<17> how cool is that.
<7> arctanx : not as cool as archimedes
<5> as cool as http://funroll-loops.org
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