Is $\displaystyle \frac{\partial}{\partial x} \ x=x\frac{\partial}{\partial x}=1$?
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2$\frac{\partial }{\partial x}x=\frac{\partial x}{\partial x}=1$. However $x\frac{\partial}{\partial x}=1$ has no sense, it is a derivative of what? – Vítězslav Štembera Dec 13 '21 at 10:48
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I'm calculating laplacian where I should express a coordinate system in terms of a and b (check my recently post) and I got $\displaystyle \frac{\partial}{\partial a} a$ which is equal to 1. OK. BUT, then I also got $\displaystyle a\frac{\partial}{\partial a}$ is it also equal to one???? – Math Lover Dec 13 '21 at 10:53
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1you should clarify in your post that the "1" is not a number but the identity operator, ie $1:f(x)\to f(x)$ – cineel Dec 14 '21 at 12:59
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No. In the way you wrote, they should be intended as operator acting on something. So the first one is $$ \partial_x(xu) = u + x\partial_x u $$ while the second is $$ x\partial_x u. $$ Note so that $\partial_x x - x \partial_x = 1$, as operators acting on something.
Angelo Zanni
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1No, since their difference is 1 (always, intended as multiplication operators). – Angelo Zanni Dec 13 '21 at 10:59
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The "No", is sus for me :/. So a better question: Are they equal? Yes or No :D, Sorry for my bad english – Math Lover Dec 13 '21 at 11:02
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1No. They are not equal. More precisely, it depends on the context, I'm intending them as multiplication operators. Example: consider the $\partial_x$ operator, then $\partial_x x =1$, but clearly the derivative operator is not equal to 1. – Angelo Zanni Dec 13 '21 at 11:07
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If you are computing the laplacian, I think $\partial_a a$ is not 1, but is an operator that act on something, so it is $\partial_a(a u) = u + a\partial_a u$, that is $\partial_ a a = a\partial_a + 1$. – Angelo Zanni Dec 13 '21 at 11:12
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check this please, https://math.stackexchange.com/questions/4330190/express-the-laplacian-partial2-x-partial2-y-in-terms-of-the-coordinate?noredirect=1#comment9031688_4330190 – Math Lover Dec 13 '21 at 11:15
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