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.
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.
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.
I think you mean operator. 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).
I’d say the $\int dx$ is the operator and the integrand is the operand.
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.
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.