>>11511220Look at the group SO(2) of rotations in the plane. These are 2x2 real orthogonal matrices. This can be viewed as a circle, picking out (1,0) as the identity, and define multiplication of points on the circle by seeing how far you have to rotate from (1,0). So (0,1)^2 = (-1,0) or (0,1)(-1,0) = (0,-1) etc.
This forms a Lie group. A smooth manifold with group structure. We look at its Lie algebra, the tangent space at the identity. The exponential map exp(x) = e^x is interesting because it takes elements of the Lie algebra and maps them to elements of the Lie group.
So we have the unit circle in the complex plane to model SO(2). Then its Lie algebra is the vertical line at (1,0), the purely imaginary line of elements ix. Then the exponential takes ix to a point on the circle, given by exp(ix) = e^ix = cosx + i sinx. You can view this geometrically as folding the line segment from (1,0) to (1,x) on the vertical line over onto the circle. This also makes it clear why it's periodic in 2pi, since if your line segment is longer than 2pi it folds back over the circle.
If you don't like complex numbers. Look at the Lie algebra of SO(2) in matrix form. Matrices in SO(2) satisfy AA^T = I. Consider a path in the Lie group A(t) with A(0) = I. Then differentiate and evaluate at 0 AA^T = 1 to get A'(0) + A^T'(0) = 0. The Lie algebra is anti-symmetric 2x2 real matrices. Up to scalar multiples there is only one, J = ((0,-1),(1,0)). Note that J^2 = -I. The set of matrices of the form aI + bJ is isomorphic to the complex numbers a + ib.
In this matrix form the formula is e^(xJ) = cosx I + sinx J
>>11511409This is what you get if you identify the Lie algebra elements with velocity.
>>11513748Identify each point (a,b) with the matrix aI+bJ and that algebra is isomorphic to a+ib.
>>11511421If you want to ignore intuition, just use this
d/dx[e^-ix(cosx+isinx)] = 0 so its constant
at 0 its 1
therefore e^ix = cosx+isinx
>>11516105Try thinking
