TRWBW | kelt: k, just double checking |

kelt | then the congagate is also a solution so... sqrt(3)/2 - i/2 blah, blah, blah... |

TRWBW | kelt: that's the same solution. |

KRUSHGROOVE | Hedos yeah I really don't know that whole example looks kinda mangled to me. |

TRWBW | kelt: each pair of conjugates gives you a pair of sin and cos solutions |

kelt | yeah TRWBW |

TRWBW | 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) 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) |

hedos | KRUSHGROOVE, 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 -) |

kelt | okay, so I need to do e^{i pi/6 + pi/3) hmm, how did you get e^{i pi/6 + pi/3) = i? |

KRUSHGROOVE | Hedos 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_she1 | is the binomial coefficient (0, 0) defined? |

kelt | e^{i pi/6 + pi/3) = e^(i pi/6) e^(pi/3)? |

TRWBW | john_she1: sure, call it 1, no harm in that |

HiLander | john_she1: yeah, it's 1 |

TRWBW | kelt: you need an x in there |

HiLander | 1 = 0!/(0!*0!) |

solis | Anyone mind looking at my problem and trying to help me integrate it? http://alnk.org/grumpyseat |

kelt | TRWBW: what do you mean? |

TRWBW | kelt: 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 |

hedos | KRUSHGROOVE, 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. |

TRWBW | kelt: 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 |

hedos | KRUSHGROOVE, 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. |

KRUSHGROOVE | hedos you could choose any real value of delta that you want... that's the whole point. But it's convenient to choose delta = 1. |

solis | When 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. |

socrates | working on uploading the pics... |

kelt | so for n = 1 we have e^[i pi/2] = cos(pi/2) + i sin(pi/2) ... uh... which is 1/2 + i/2 ? |

socrates | basically 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 |

kelt | hmm, I don't remember my sin's and cos's lol |

socrates | and I'm not quite understanding it |

kelt | e^{i (pi/6 + pi/3)} = e^{i pi/2) |

TRWBW | kelt: f'+f=0 solution is e^((i*pi/2)*x)=e^(-x) |

socrates | http://farm1.static.flickr.com/193/445671091_49b717d7a8_m.jpg picture 1 |