4 ms·
e^(epsilon) ~ 1 + epsilon in general, because of the definition of the derivative and the fact that d/dx (e^x) = e^x
by ThrustVectoring 10y ago
e^(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.