Comment on UwU brat mathematician behavior

• Zagorath@aussie.zone ⁨3⁩ ⁨months⁩ ago

  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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  • Chrobin@discuss.tchncs.de ⁨3⁩ ⁨months⁩ ago

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

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    • Zagorath@aussie.zone ⁨3⁩ ⁨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.

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      • Chrobin@discuss.tchncs.de ⁨3⁩ ⁨months⁩ 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.

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