#math - Wed 4 Apr 2007 between 01:23 and 01:35

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TRWBWkelt: k, just double checking
keltthen the congagate is also a solution
so... sqrt(3)/2 - i/2
blah, blah, blah...
TRWBWkelt: that's the same solution.
KRUSHGROOVEHedos yeah I really don't know that whole example looks kinda mangled to me.
TRWBWkelt: each pair of conjugates gives you a pair of sin and cos solutions
keltyeah TRWBW
TRWBWkelt: the basic idea here is that the functions that can be written as a*e^ix+b*e^ix are exacty the same as the set of functions that can be written a*cos(x)+b*sin(x)
oops
kelt: the basic idea here is that the functions that can be written as a*e^ix+b*e^-ix are exacty the same as the set of functions that can be written a*cos(x)+b*sin(x)
hedosKRUSHGROOVE, how would you have done it? Say, when you get to the |x-1||x+3| < Epsilon part. What should be done?
TRWBW(left out a -)
keltokay, so I need to do e^{i pi/6 + pi/3)
hmm, how did you get e^{i pi/6 + pi/3) = i?
KRUSHGROOVEHedos ok, that line you're talking about is just saying that if delta = 1, then x is in the interval [0,2] since |x-1| < delta => (delta) -1 < x < (delta) + 1 => 1-1 < x < 1+1 => 0<x<2
john_she1is the binomial coefficient (0, 0) defined?
kelte^{i pi/6 + pi/3) = e^(i pi/6) e^(pi/3)?
TRWBWjohn_she1: sure, call it 1, no harm in that
HiLanderjohn_she1: yeah, it's 1
TRWBWkelt: you need an x in there
HiLander1 = 0!/(0!*0!)
solisAnyone mind looking at my problem and trying to help me integrate it? http://alnk.org/grumpyseat
keltTRWBW: what do you mean?
TRWBWkelt: remember the solutions are e^(rx) where r solves r^6+1=0. that's easy to see, f(x)=e^(r*x), f^(6)(x)=r^6f(x), r^6f(x)+f(x)=f(x)(r^6+1)=0
hedosKRUSHGROOVE, I understand that. I wonder how that is a proof. How taking a random value of delta which gives me the interval of X a demonstration. That's what bugs me about those limits demonstration, it seems like you can do anything.
TRWBWkelt: that is your solutions are of the form f(x)=e^(e^(pi*i/6+n*pi*i/3)*x) and you are combining pairs to get sin's and cos's
hedosKRUSHGROOVE, so we took the largest value of x for a delta = 1, that is x=2. I suppose there's a reason for that and why it is valide and why it makes the demonstration valid.
KRUSHGROOVEhedos you could choose any real value of delta that you want... that's the whole point. But it's convenient to choose delta = 1.
solisWhen I integrate http://alnk.org/grumpyseat I get A1^2*(-a/4)*(2*r^2+2a*r+a^2)*e^(-2r/a) evaluated from 0 to infinity.
socratesworking on uploading the pics...
keltso for n = 1 we have e^[i pi/2] = cos(pi/2) + i sin(pi/2) ... uh... which is 1/2 + i/2 ?
socratesbasically the deal is the instructor sucks.... I have a major quiz tomorrow and he pretty much flat out told us that we would see all these circles combined into one huge circle
and that would be our test
kelthmm, I don't remember my sin's and cos's lol
socratesand I'm not quite understanding it
kelte^{i (pi/6 + pi/3)} = e^{i pi/2)
TRWBWkelt: f'+f=0 solution is e^((i*pi/2)*x)=e^(-x)
socrateshttp://farm1.static.flickr.com/193/445671091_49b717d7a8_m.jpg
picture 1

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