latex align command

[topsquark]

In $2x-\frac{\text{d}}{\text{dx}}$ it looks like function part is missing, looks like subtracting operator from 2x.

Think of this as
$2x \dfrac{d^{\,0}}{dx^{\,0}} - \dfrac{d}{dx}$

The operator $d^{\,0}/dx^{\,0}$ is the "identity derivative operator," that is, it tells us to just leave what it's operating on alone.

So
$\left ( 2x \dfrac{d^{\,0}}{dx^{\,0}} - \dfrac{d}{dx} \right ) (3x^2) = 2x \cdot \dfrac{d^{\,0}}{dx^{\,0}}(3x^2) - \dfrac{d}{dx} (3x^2) = 6x^3 - 6x$

-Dan

 

[Bruce]

This site handles it fine when you Reply to an entire message, but messes up when you select a part and Reply to that

That's a good tip

 

[Bruce]

So
$\left ( 2x \dfrac{d^{\,0}}{dx^{\,0}} - \dfrac{d}{dx} \right ) (3x^2) = 2x \cdot \dfrac{d^{\,0}}{dx^{\,0}}(3x^2) - \dfrac{d}{dx} (3x^2) = 6x^3 - 6x$

If I get Dr. Peterson's thinking, we say d/dx by itself means some function in general or still unknown to be differentiated later, and after some algebra steps the d/dx ends up next to a function and we change meaning to operator.

So I'm correct to say your first $\frac{d^{\,0}}{dx^{\,0}} \text{ and } \frac{d}{dx}$ are like parameters, meaning some function to differentiate later in process? That makes it ok to use distribution and now see them as operators in front of a function to differentiate (except we don't do differentiation with identity derivative operator, we ignore it.)

 

[Dr.Peterson]

If I get Dr. Peterson's thinking, we say d/dx by itself means some function in general or still unknown to be differentiated later, and after some algebra steps the d/dx ends up next to a function and we change meaning to operator.

So I'm correct to say your first $\frac{d^{\,0}}{dx^{\,0}} \text{ and } \frac{d}{dx}$ are like parameters, meaning some function to differentiate later in process? Then it is valid to use distribution and get operators in front of the function to differentiate (except we don't differentiate with the identity derivative operator, we ignore it.

No, I wouldn't compare them to parameters, which are constants that can be modified. We don't modify $\frac{d}{dx}$.

A better comparison would be to functions, which is essentially what operators are. Given functions f and g, we can talk about a function f+g. We can't evaluate this until we give it an input (argument), and then we have $(f+g)(x)=f(x)+g(x)$.

Similarly, we can write $\left(2x-\frac{d}{dx}\right)$, which is the sum of two operators (functions of functions), but does nothing until we apply it: $$\left(2x-\frac{d}{dx}\right)f(x)=2xf(x)-\frac{d}{dx}f(x).$$

 

[Bruce]

A better comparison would be to functions, which is essentially what operators are. Given functions f and g, we can talk about a function f+g. We can't evaluate this until we give it an input

Ok, that helps more, thank you.