I thought it looked like "imaginary i", but it doesn't make sense in that equation to me. How does e^(i*pi) = -1 work? I need some derivations..
ty, now just reading over the proof for euler's to finish jogging my memory. Ho hum, I used to be a much better nerd. I can barely remember how to integrate. Sx = (x^2)/2 ?
^^^ lol, old but classic. FYI, here's a proof of euler's. http://en.wikipedia.org/wiki/Euler's_formula Define the function g(x) by g(x) \ \stackrel{\mathrm{def}}{=}\ e^{ix} .\ Considering that i is constant, the first and second derivatives of g(x) are g'(x) = i e^{ix} \ g''(x) = i^2 e^{ix} = -e^{ix} \ because i 2 = −1 by definition. From this the following 2nd-order linear ordinary differential equation is constructed: g''(x) = -g(x) \ or g''(x) + g(x) = 0. \ Being a 2nd-order differential equation, there are two linearly independent solutions that satisfy it: g_1(x) = \cos(x) \ g_2(x) = \sin(x). \ Both cos(x) and sin(x) are real functions in which the 2nd derivative is identical to the negative of that function. Any linear combination of solutions to a homogeneous differential equation is also a solution. Then, in general, the solution to the differential equation is g(x)\, = A g_1(x) + B g_2(x) \ = A \cos(x) + B \sin(x) \ for any constants A and B. But not all values of these two constants satisfy the known initial conditions for g(x): g(0) = e^{i0} = 1 \ g'(0) = i e^{i0} = i \ . However these same initial conditions (applied to the general solution) are g(0) = A \cos(0) + B \sin(0) = A \ g'(0) = -A \sin(0) + B \cos(0) = B \ resulting in g(0) = A = 1 \ g'(0) = B = i \ and, finally, g(x) \ \stackrel{\mathrm{def}}{=}\ e^{ix} = \cos(x) + i \sin(x). \ Q.E.D.
That won't necessarily help, it's over my head too... I'm about to finish grad school and I haven't taken a math class since AP Calculus my senior year of H.S.
um...wow. i feel like i have some fuc$ing down syndrome myself, right about now. hey at least i can add my bills up right?