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UwU brat mathematician behavior (mander.xyz)
submitted 1 year ago by to c/science_memes@mander.xyz
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[–] 1 point 1 year ago (5 children)

I think you mean operator. The operand is the target of an operator.

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  • [–] 2 points 1 year ago (4 children)

    The 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).

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  • [–] 2 points 1 year ago (3 children)

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

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  • [–] 1 point 1 year ago (2 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.

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  • [–] 2 points 1 year ago (1 child)

    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.

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  • [–] 0 points 1 year 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.

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