4 ms·
Starting for Euler's formula you can write cos(x) = (exp(ix) + exp(-ix))/2 so the cosine is in some sense a "type of exponential". If the cosine is a type of ex
by eigenket 6y ago
Starting for Euler's formula you can write cos(x) = (exp(ix) + exp(-ix))/2 so the cosine is in some sense a "type of exponential". If the cosine is a type of exponential then its inverse (the function that takes you to the angle measurement) is inherently a "type of logarithm".
This is a very handwavey statement. I would say a more precise statement is that the proper exponential and log maps are the "correct" thing to use to map multiplication to adidtion and vice-versa. They are exactly the maps that do this properly. The fact that cosine is a sum of exponentials means that you can write the inverse of cosine as a (slightly ugly) formula in terms of logarithms (arcos(z) = -i log(sqrt(1-z^2) + z). This is the sense in which the map to angle measures is a "type of log".
- jacobolus 6y agoThe angle measure i acos(x) is the logarithm of the rotation x + i√(1 − x²). If you want to compute the product of two rotations, you can take logarithms and rewrite it as a sum of angle measures: [x + i√(1 − x²)][y + i√(1 − y²)] = exp(i acos(x) + i acos(y)) [x + i√(1 − x²)][y − i√(1 − y²)] = exp(i acos(x) − i acos(y)) Now if you just look at the real parts, you can write: xy − √(1 − x²)√(1 − y²) = cos(acos(x) + acos(y)) xy + √(1 − x²)√(1 − y²) = cos(acos(x) − acos(y)) Taking the sum on each side: 2xy = cos(acos(x) + acos(y)) + cos(acos(x) − acos(y)) This method is rather more cumbersome (due to rescaling and then converting the factors to rotations as an intermediate step) but fundamentally based on the same concept as: xy = exp(log(x) + log(y))