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Comments:
<0> when is the next seminaire ?
<1> seminaire?
<0> seminar
<1> What seminar?
<0> on efnet its like a common thing
<0> people all agree upon a date,and someone is to give a seminar,from the channels "elders"
<0> and all listen and pose questions at the end
<0> and the "elder" is to answer and stuff
<0> its pretty cool
<2> hasn't happened in a while methinks
<0> except im banned from efnet so i would have guessed they're done here too
<1> how did you get banned?
<3> spx2: well register a channel and try to get something started
<0> __mikem: its not relevant
<4> seminars are wheenever someone volunteers (or otherwise) to do one
<0> have there been any up until now ?
<4> yes, but not for a while
<4> afik the efnet ones are the same, relayed from #mathematics
<2> does the one point compactification of R have a special name?
<5> S^1
<2> yeah, but I was wondering if it had a real name
<2> like that of C is the riemann sphere
<5> that's quite real
<2> perhaps the riemann circle or something
<5> uh, "the circle"
<6> What is the one point compactification?
<2> koro: shrug, alright
<7> Alexandroff
<5> add a point to the space, and extend the topology so that it becomes compact with that additional point
<6> and what point do you add to R?
<8> Hi again!
<5> just *a* point
<3> xerox: it's ajust a point, but you can call it "infinity" if you like. or "betty"
<5> i.e. you consider the space R union {p}, where {p} is a one-element set with p not in R
<6> aka the projective line ?
<8> How can I get the b in "d = sqrt(a^2 + b^2)" to be alone on one side of the equation? I've problems on how to get it out from under its square root...
<4> rapha: if it was d=sqrt(b) what would you do?
<6> oh, now I'm curious, what do you add to C?
<3> d^2=a^2+b^2 help?
<5> xerox: just a point!
<3> xero: "betty" again
<3> xerox: betty gets around a lot, nudge nudge, wink wink
<5> no, betty is for R, for C it's johnny
<8> _llll_: I've no idea :( ... maybe take the square root... would that be sqrt(d) = b then?
<4> xerox: it doesnt matter what space it is, you just do 'X [disjoint union] {*}' for any 1-point space {*}, and then define a topology
<8> TRWBW: for the problem at hand yes, but not for learning on how to do it myself :-{
<4> rapha: you'd square both sides, and get sqrt(d)=b. so try that approach on your problem
<6> a projective C is just the projective version of R or something more?
<6> *R^2
<5> what do you mean by "projective C" and "projective version of R"?
<8> _llll_: sqrt(d) = a^2 + b^2; sqrt(d)/a^2 = b^2; sqrt(sqrt(d)/a^2) = b ?
<3> xerox: the one point construction, and someone check me on this, is that you take all bad open sets O of S, such that S-O isn't compact, and add a new open set O'=O+{"delores"}. i think
<9> is there anyway to factor x^4 + 1?
<10> hi. has anyone here sat STEP exams?
<8> Can somebody tell me if http://www.mathbin.net/7298 is correct (and if yes, how it can be further simplified?)
<3> f3dt: do you know the connection between root finding (finding zeros) and factoring?
<4> rapha: sqrt(d) = a^2 + b^2 is right, but now what did you do?
<8> _llll_: I re-did it in http://www.mathbin.net/7298
<3> lll: didn't he start with d=sqrt(a^2+b^2)?
<9> yeah...so it won't factor
<9> i forgot
<8> TRWBW: yes
<3> f3dt: well it depends, are you doing it on the reals or the complex
<9> reals
<9> is there a way to write 1/(x^4+1) in partial fractions?
<3> f3dt: then you have your answer. x^4 is always >=0, so its >=1
<4> rapha: i cant understand that link
<3> f3dt: yes, but you need complex numbers
<9> gar
<11> f3dt: x^4+1 is irreducible over the reals
<11> yes, what trwbw said
<11> I am too slow
<9> then i have no clue how to integrate that
<3> f3dt: maybe you should bite the bullet and go complex
<8> _llll_: I tried to solve it to b... did I type it wrongly?
<11> Definite or indefinite integral?
<4> rapha: i couldnt make any sense out of that page, it was just a mess of symbols
<9> indefinite
<11> nuts
<3> d=sqrt(a^2+b^2)? -> d^2=a^2+b^2 -> b^2=a^2-d^2 -> ?
<8> _llll_: wierd, it is a MathBin page for me, with some stuff rendered by LaTeX
<11> Well, we can always ask Mathematica
<11> Cover your eyes, everyone; mbot doesn't work in private :(
<4> rapha: yes, but what you wrote on it is not understandable
<11> % Integrate[1/(x^4+1), x]
<11> Nor is it even in the room
<12> This algorithm http://rafb.net/p/AggVCZ94.html calculates 5 digits of pi by iterating over a list of 5 integers 13 times. the list starts as 3,0,0,0,0 (ie 3.0000), increases to 3.7738, then decreases to 3.1415. Output here http://rafb.net/p/3p4XxM29.txt . It works for any number of digits given enough iterations, and never increases past 4.0. Can anyone tell me the name of that algorithm?
<13> f3dt: did you try a trig subutition (my initial guess)
<9> ha
<4> id guess you cant find an anti-derivative
<13> I found the integral in maple
<11> Well, f3dt, this is happy to give you an answer http://integrals.wolfram.com/index.jsp
<8> _llll_: it is meant to be three lines, the upper two have a | behind which I say what I'll do to get to the next line... we learned to do it like that in school
<11> And it's what mbot would have given you if it had been in the room
<13> f3dt: so my guess is still trig subutition
<8> TRWBW: so basically you're saying "pretend the sqrt wasn't there"?
<4> rapha: that is a very bad way to present things, your school should be ashamed
<8> Hmm
<9> what would i substitute for that?
<14> f3dt - it is guaranteed to be integrable (i.e. antidifferentiable) by the method of partial fractions.
<11> And I don't know how it got what it got, but it smells of trig substitution
<8> _llll_: how would you write it down?
<14> _llll_ - then you are forgetting a fundamental factorisation theorem...
<8> I want to learn to write it down so that mathematicians like you can understand it!
<9> but i have to use complexes if i use partial fractions?
<14> of course not.
<3> f3dt: it's the direct way. there may be a trick i don't know.
<4> rapha: it is usually better to write in words what you are doing. of course an experience person would probably miss out all the intermediate steps there...
<4> rapha: the second line is ok, but applying the square root does not give what you have on the third line
<8> Oh! I made a mistake!
<13> f3dt: I have the answer from maple if your intersted, but my suggestion will either big a trig subutition, but I do belive there is a a^4+b^4 facorization, but not sure, not somethign I see or use commonally
<3> somiaj: not over the reals. let me ask you this, how could x^4+1=0 with x real?
<3> somiaj: actually that's not a proof, it could factor as quadradics, but it doesn't
<13> TRWBW: ahh, yea, I know there are four roots, I was thinking they were the four on the axis, but the are the four roots that would be on the lines y=x and y=-x on the compex plain now that I think about it
<3> somiaj: um, maybe it does, let me reconsider
<8> _llll_: is it better like this: http://www.mathbin.net/7302 ?
<4> much better
<4> i dont like the |-a^2 stuff, but once you know what it means, it is at least understandable
<13> TRWBW: because the roots would be 1e^(i * theta) where theta = pi/4, 3pi/4, 5pi/4 and 7pi/4 if my mind is working at all
<8> _llll_: how would the "| ..." stuff normally be written? "subtract a^2"?
<3> right so you have complex conjugates you should be able to multiply
<13> TRWBW: oh yea, so multipling the complex congets should give two real factors with x^2 involved....
<4> i'd just write "subtracting a^2 gives" between the two equations or something. (well, probably i'd just write down the answer and miss out all the steps, but that isnt good advice ;))
<13> well maple doesn't factor x^4+1....
<3> (x-exp(ipi/4))(x-exp(i5pi/4))=x^2-(exp(ipi/4)+exp(i5pi/4))x+exp(i6pi/4)
<14> what is (x^2 - sqrt(2) x + 1)(x^2 + sqrt(2) x + 1)?
<3> okay so any polynomial is reducible over the reals to a product of quadratics and linears then?
<3> that sounds right, like something i had forgotten
<14> of course.
<8> _llll_: Well, I'm on the niveau of a 4th-cl***-student now, so I'm happy for all the intermediate steps I can get :-)
<3> trivial when you say it out loud.
<8> _llll_: can you understand the following and tell if I did it correctly? http://www.mathbin.net/7303
<4> well the roots come in conjugate pairs. i still wouldnt guess you could integrate it, even from that
<9> how did you get thos roots though?
<3> that makes it pretty straightforward, just a1x+b1 over each quadratic factor, the numerator will be a cubic, and with 4 variables you should be able to solve for a numerator of 1
<3> i mean after you add them
<13> the answer apears to have a sqrt(2) x term in it, so I am inclined to think that is the correct method (tis nice to know the answer your working twoards....ie I cheat)
<4> roots of x^4+1 are pretty seasy to find, just use the polar form of a complex number
<15> trwbw: http://blog.plover.com/math/factoring.html discusses the factorization of x^4+1 over the reals.
<16> via reddit: http://www.robertwechsler.com.nyud.net:8080/images/applied_geometry.jpg
<15> x^4 + 1 = (x^2 - x + 1)(x^2 + x + 1)
<3> k, i saw it pretty much right after i said "not you can't" with "um,..., let me reconsider"
<8> Putting those carts together was a hell of a lot of work ivan` :-)
<9> erm..what's that before 2x? it doesn't display on my computer
<3> yrlnry: sweet
<15> f3dt: square root sign
<9> oh
<3> (x^2+1 +- sqrt(2)x)^2=x^4+2x^2+1-2x^2=x^4+1
<15> x^4 + 1 = (x^2 - sqrt(2)x + 1)(x^2 + sqrt(2)x + 1)
<3> oops wrote that wrong
<15> That is one heck of an abuse of notation.
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