4 ms·

I think what is confusing about differentials is that variables are meta-objects. You can apply function to a number, you can apply function to another function

I think what is confusing about differentials is that variables are meta-objects. You can apply function to a number, you can apply function to another function and so on but you cannot really apply function to a variable, you apply it to something that the variable stands for. The introduction of calculus in my university went like this: At first we introduce derivative f’(x) as a limit [f(x+h)-f(x)]/h. It is a linear function. We call d the function that takes f(x) and h and returns f’(x)*h (or it can be defined by currying, whatever). So now we can write df and it makes sense. Ok, so far so good. But what is even dx? We cannot just apply a function to a meta-symbol. Variables are not first-level objects, they are just notational fiction. You may say that it is an identity function, but such explanation breaks down when more free variables are introduced.

codethief 4y ago
You're making a good point. However, taking inspiration from differential topology (and differential forms in particular), the notation dx can be made sense of once you interpret x as a coordinate function x: M -> R on a manifold. So in a sense you are replacing the variable x with a function x (whose parameter is usually left unspecified).
- LudwigNagasena 4y ago
  Yeah, it’s the standard way to do it. But there is a reason the 7th point is called “stay away from differentials.” It’s cryptic and unintuitive for many people. I wonder whether name binding concepts from CS theory a la lambda calculus can be leveraged for an alternative formalization.
  - selimthegrim 4y ago
    I think usually slightly alternate definition of derivative is used - https://en.wikipedia.org/wiki/Total_derivative https://en.wikipedia.org/wiki/Total_derivative
twos222 4y ago
If f is a function and x is a variable then dx means that you should express f with respect to an input x, so if A is the area of a a circle then what is dA? since A = pir^2 you have that dA = 2pirdr since A = pi(l/(2pi))^2 where l is the perimeter then you have dA = pi2(l/(2pi))(1/(2pi))dl, so the idea of dA is something related to the diferential of the variable used to define A. The use of the chain rule allow us to operate with differentials. For example, since here r = l/(2pi) we have that dr = dl/(2pi) so that dA = 2pirdr = 2pi(l/(2pi))dl/(2pi) = 1/(2pi)ldl. An finally this implies that dA/dl = 1/(2pi)l
- ericbarrett 4y ago
  HN formatting has swallowed your asterisks; try writing them as \*
aj7 4y ago
It’s my experience that there are some people who cannot understand calculus, no matter how hard they try. The fact that they do try is often indicative of considerable talents in other areas.

codethief 4y ago

You're making a good point. However, taking inspiration from differential topology (and differential forms in particular), the notation dx can be made sense of once you interpret x as a coordinate function x: M -> R on a manifold. So in a sense you are replacing the variable x with a function x (whose parameter is usually left unspecified).

LudwigNagasena 4y ago
Yeah, it’s the standard way to do it. But there is a reason the 7th point is called “stay away from differentials.” It’s cryptic and unintuitive for many people. I wonder whether name binding concepts from CS theory a la lambda calculus can be leveraged for an alternative formalization.
- selimthegrim 4y ago
  I think usually slightly alternate definition of derivative is used - https://en.wikipedia.org/wiki/Total_derivative https://en.wikipedia.org/wiki/Total_derivative

twos222 4y ago

If f is a function and x is a variable then dx means that you should express f with respect to an input x, so if A is the area of a a circle then what is dA? since A = pir^2 you have that dA = 2pirdr since A = pi(l/(2pi))^2 where l is the perimeter then you have dA = pi2(l/(2pi))(1/(2pi))dl, so the idea of dA is something related to the diferential of the variable used to define A. The use of the chain rule allow us to operate with differentials. For example, since here r = l/(2pi) we have that dr = dl/(2pi) so that dA = 2pirdr = 2pi(l/(2pi))dl/(2pi) = 1/(2pi)ldl. An finally this implies that dA/dl = 1/(2pi)l

ericbarrett 4y ago
HN formatting has swallowed your asterisks; try writing them as \*

aj7 4y ago

It’s my experience that there are some people who cannot understand calculus, no matter how hard they try. The fact that they do try is often indicative of considerable talents in other areas.