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 I hate long decimals, I just don't feel right putting a rounded answer on my paper ;_; <3 the pi/e/i/etc. buttons on the calculator n.n 
05-07-2008, 04:26 AM
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| ^Tbh, I actually prefer decimals. Im a physicist so I dont mind rounding. Its only those neeky mathematicians who say pi/3 instead of 1.05
05-07-2008, 06:57 PM
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| Because decimals are, most of the time, approximations, and not exact values of a given situation. Decimals also introduce more room for potential errors; when exact values are mistaken, usually the answer goes from correct to really wrong really fast, and is a form of consistency for our theoretical, perfect little worlds we have in our heads. Also, we hate significant figures. To the death.
05-07-2008, 07:12 PM
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| ugh...when I enter a science classroom it's just 2+2=5 for me all of the sudden ;_;
05-07-2008, 11:20 PM
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| You should do your own homework. But I will do it for you this once. The easiest method is to use DeMoivre's Theorem, which presumably you know, exp(iy) = (cos(y + 2(PI)n) + i sin(y + 2(PI)n)). n is an integer. put x^3 = a*exp(iy). So y = 0 and a = 192/81. Then x = (a^(1/3))*exp(i(y/3)) x = (a^(1/3))*(cos(y/3 + 2(PI)n/3) + i sin(y/3 + 2(PI)n/3)) Here we have y = 0 so x = (a^(1/3))*(cos(2(PI)n/3) + i sin(2(PI)n/3)) Then stick in integers for n, n = 0 x = (a^(1/3))*(cos(0) + i sin(0)) x = (a^(1/3)) = 4/3 n = 1 x = (a^(1/3))*(cos(2(PI)/3) + i sin(2(PI)/3)) x = (2/3)*(1 + i Sqrt(3)) n = 2 x = (a^(1/3))*(cos(4(PI)/3) + i sin(4(PI)/3)) x = (2/3)*(1 - i Sqrt(3)) n >= 3 just cycles through these solutions. These solutions define an equilateral triangle in the complex plane. PuPu did it a really nice way, and yeah you have to get 3 answers.
05-13-2008, 04:56 PM
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