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UwU brat mathematician behavior (lemmynsfw.com)

submitted 6 days ago by [email protected] to c/[email protected]

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[–] [email protected] 2 points 5 days ago (1 children)

I'd say the $\int dx$ is the operator and the integrand is the operand.

[–] [email protected] 1 points 5 days ago (1 children)

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.

[–] [email protected] 2 points 5 days ago (1 children)

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.

[–] [email protected] 1 points 1 day ago

Ok that's some really interesting context I didn't know. I've only ever seen it done the mathematician's way with dx at the end. Learning physicists do it differently explains why the person in the post would want to discuss moving it around.

But I still think they have to mean "if dx can be treated as an operand". Because "if dx can be treated as an operator" doesn't make sense. It is an operator; there's no need to comment on something being what it objectively is, and even less reason to pretend OOP's partner was angry at this idea.