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

### NY Lost Funds

 TRWBW kelt: k, just double checking kelt then the congagate is also a solutionso... sqrt(3)/2 - i/2blah, 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)oopskelt: 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