2010-05-23, 04:33 PM
Funny thing you mention it, actually. Me and my friend have a habit of taking complicated mathematical formulas and telling them repeatedly to many people in order to annoy them. Over the years we have passed many mathematical subjects, because the guys we were trying to annoy actually learned these subjects in math class. We have already gone through trigonometry, calculus, infinity maths, complex numbers and just a few weeks ago we started using complex numbers with euler's formula (e^ai = cos(a) + isin(a)). Allow me to explain:
e^pi*i = cos(pi) + i*sin(pi) = -1 + 0
e^pi*i + 1 = 0 (according to wikipedia, this formula was voted the most beautiful mathematical formula ever for using all 5 of math's most important constants)
e^i*(pi/2) = cos(pi/2) + i*sin(pi/2) = i
therefore:
i^i = e^i*(pi/2)^i = e^(i^2)*(pi/2) = e^-pi/2
i'th root of i = i^1/i = e^i*(pi/2)^(1/i) = e^(pi/2)
(-1)^-i = i^2^-i = (e^i*(pi/2))^2^-i = e^i*pi^-i = e^-(i^2)*pi = e^pi (also known as gelfond's constant)
We usually tell the above paragraph at least two times a day to random students. Yes, I am quite the annoying type of a person.
e^pi*i = cos(pi) + i*sin(pi) = -1 + 0
e^pi*i + 1 = 0 (according to wikipedia, this formula was voted the most beautiful mathematical formula ever for using all 5 of math's most important constants)
e^i*(pi/2) = cos(pi/2) + i*sin(pi/2) = i
therefore:
i^i = e^i*(pi/2)^i = e^(i^2)*(pi/2) = e^-pi/2
i'th root of i = i^1/i = e^i*(pi/2)^(1/i) = e^(pi/2)
(-1)^-i = i^2^-i = (e^i*(pi/2))^2^-i = e^i*pi^-i = e^-(i^2)*pi = e^pi (also known as gelfond's constant)
We usually tell the above paragraph at least two times a day to random students. Yes, I am quite the annoying type of a person.