State the conditions that partial derivatives commute, namely, $D_1D_2f = D_2D_1f$.
I understand how to prove that these partial derivatives are equal but I don't understand what commute means. Please help.
State the conditions that partial derivatives commute, namely, $D_1D_2f = D_2D_1f$.
I understand how to prove that these partial derivatives are equal but I don't understand what commute means. Please help.
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Technically, you can think of $D_1$ and $D_2$ as operations on the function $f$. For the operations $D_1$ and $D_2$ to commute we need that after applying $D_1 D_2$ to $f$ we get $D_2 D_1 f$ or $$ D_1 D_2 f(x_1, x_2) = D_2 D_1 f(x_1, x_2) \Leftrightarrow D_1 D_2 f - D_2 D_1 f(x_1, x_2) = 0 \Leftrightarrow (D_1 D_2 - D_2 D_1)f(x_1, x_2) = 0. $$ Yes that last bit means that $D_1 D_2 - D_2 D_1$ is also an operation on $f(x_1, x_2)$ and it will be the zeroing operation. As an exercise you may want to try to rewrite the above tautology in terms of a simple function, say $\exp\{ - (x^2 + y^2) \}.$
We can think of more general operations like, say multiplying $f(x_1,x_2)$ by $x_1$, call this operation $M_1$. That is, $$ M_1 f(x_1, x_2) = x_1 f(x_1, x_2)$$ a new function. Then we see that (this is an exercise) $M_1 D_2 f = D_2 M_1 f$ but $M_1 D_1 f \neq D_1 M_1 f$. In fact $$( M_1 D_1 - D_1 M_1) f = 1 $$ the constant function.
I know this is extra information but I just wanted to point out that this formalism is not arbitrary. It turns out that the certain operations (like differentiation and multiplication by a variable) have a nice algebraic structure and that commutativity is usually not part of it. However, the commutator substructure can be built by considering the bits left over: the operations $$ AB - BA$$ as above. The properties of this substructure are important to many branches of engineering and physics. The example above (with a small but imaginative modification) can be used to describe the uncertainty of position and momentum observations according to the laws of quantum mechanics.
Commute just means that the order of derivatives that we take does not matter. One could take the partials with respect to $x$ and then the partial with respect to $y$ or instead the partial to $y$ and then $x$. The final answer is independent of the order in which we take our derivatives.