Comment on UwU brat mathematician behavior
Zagorath@aussie.zone 3 days agoThe operand is the target of an operator
Correct. Thus, dx is an operand. It’s a thing by which you multiply the rest of the equation (or, in the case of dy/dx, by which you divide the dy).
Chrobin@discuss.tchncs.de 3 days ago
I’d say the $\int dx$ is the operator and the integrand is the operand.
Zagorath@aussie.zone 3 days ago
You’re misunderstanding the post. Yes, the reality of maths is that the integral is an operator. But the post talks about how “dx can be treated as an [operand]”. And this is true, in many (but not all) circumstances.
∫(dy/dx)dx = ∫dy = y
Or the chain rule:
(dz/dy)(dy/dx) = dz/dx
In both of these cases, dx or dy behave like operands, since we can “cancel” them through division. This isn’t rigorous maths, but it’s a frequently-useful shorthand.
Chrobin@discuss.tchncs.de 2 days ago
I do understand it differently, but I don’t think I misunderstood. I think what they meant is the physicist notation I’m (as a physicist) all too familiar with:
∫ f(x) dx = ∫ dx f(x)
In this case, because f(x) is the operand and ∫ dx the operator, it’s still uniquely defined.