3 ms·
No, it makes sense. exp(i epsilon) = 1 + i epsilon, to first order, i.e. when epsilon is near 0. You can think of 1 + i epsilon as a tiny rotation, when you mul
by abecedarius 10y ago
No, it makes sense. exp(i epsilon) = 1 + i epsilon, to first order, i.e. when epsilon is near 0. You can think of 1 + i epsilon as a tiny rotation, when you multiply by it. exp(i t) = the product of a lot of tiny rotations, just as in real numbers exp(x) = the product of a lot of tiny scalings. The product of rotations equals one rotation by the sum of their (individually tiny) angles. As t increases continuously, exp(i t) rotates around from 1 continuously. The unchanging speed of rotation corresponds to how in e.g. bacterial growth, exp(t), the population gets some constant factor bigger in every equal interval.
(I first got this picture from the Feynman lecture on algebra, http://www.feynmanlectures.caltech.edu/I_22.html http://www.feynmanlectures.caltech.edu/I_22.html, which doesn't seem to ever explain it that way explicitly. It's funny how memory works.)
- jordigh 10y agoOkay, this makes a little more sense to me, since it's just the power series again. But I don't think that calling a rotation a growth makes sense. Also, to me, the approximation e^(i epsilon) ~ 1 + i epsilon kind of begs the question.
- ThrustVectoring 10y agoe^(epsilon) ~ 1 + epsilon in general, because of the definition of the derivative and the fact that d/dx (e^x) = e^x
- jordigh 10y agoRight, the latter fact is equivalent to considering exp the solution of a differential equation, which is a much bigger fact. Treating the approximation as an innocent brute fact is what I considered begging the question, because why should this differential equation be valid for complex values?
- abecedarius 10y agoI was trying to make a geometric intuition accessible to high schoolers who haven't mastered calculus -- though I haven't tried it on an actual teenager. Of course there's also value in looking deeper into the foundations.