Comment on UwU brat mathematician behavior
Zagorath@aussie.zone 2 days ago“operative” instead of, uh, something else
I think they meant “operand”. As in, in the way dy/dx can sometimes be treated as a fraction and dx treated as a value.
Comment on UwU brat mathematician behavior
Zagorath@aussie.zone 2 days ago“operative” instead of, uh, something else
I think they meant “operand”. As in, in the way dy/dx can sometimes be treated as a fraction and dx treated as a value.
Chrobin@discuss.tchncs.de 2 days ago
I think you mean operator. The operand is the target of an operator.
Zagorath@aussie.zone 2 days ago
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 2 days ago
I’d say the $\int dx$ is the operator and the integrand is the operand.
Zagorath@aussie.zone 2 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.