Comment on UwU brat mathematician behavior
Zagorath@aussie.zone 10 months 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 10 months 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 10 months ago
I think you mean operator. The operand is the target of an operator.
Zagorath@aussie.zone 10 months 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 10 months ago
I’d say the $\int dx$ is the operator and the integrand is the operand.
Zagorath@aussie.zone 10 months 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.