3 ms·

The problem is that d/dx(dy/dx) is not d^2 y/(dx)^2. d/dx (dy/dx) is really d(dy/dx)/dx, which is ((d^2 y)/dx + dy d(1/dx))/dx = d^2 y/dx - dy/dx (d^2 x)/(dx)^

The problem is that d/dx(dy/dx) is not d^2 y/(dx)^2. d/dx (dy/dx) is really d(dy/dx)/dx, which is ((d^2 y)/dx + dy d(1/dx))/dx = d^2 y/dx - dy/dx (d^2 x)/(dx)^2, where df is the actual differential.

bikenaga 3y ago
We're probably talking about different things. I'm not using "d/dx" as the differential operator "d" divided by the differential "dx", which is what I think you mean. I mean the compound symbol "d/dx" from calculus, which means differentiation with respect to x - or alternatively (from a differential geometric point of view) the tangent vector "d/dx" operating on the function f. I meant to explain why the notation "d^2 f/dx^2" is the way it is. It looks funny, and students would often ask why the 2's were "in different places". But it makes sense when you remember that "d/dx" is the derivative operator.
- impossiblefork 3y ago
  I am talking about d/dx as the differential operator d divided by dx. That's the way to make it actually work, algebraically. d^2 f/dx^2 is actually wrong as the way to write the second derivative unless df/dx = 0 or d^2 x/(dx)^2 = 0; the first case is trivial, and the second is almost never the case. Suppose for example that in actually x = t^2. Then dx = 2t dt, d(dx) = 2(dt)^2 + 2 t (d^2 t) and we get (2(dt)^2 + 2t(d^2 t))/(2t dt)^2 = 0.5 + (...) d^2 t/(dt)^2, so it certainly isn't zero. But if you have actual differentials you can do algebra.
  - bikenaga 3y ago
    > I am talking about d/dx as the differential operator d divided by dx. Okay, that's what I figured from your earlier computation. (One note: If "d" means exterior differentiation, then d^2 = 0.) But if you mean d/dx in this way you should make sure you tell people, because that's not the interpretation it has in calculus (say a typical calc class in a typical college) or in differential geometry. In those cases, d/dx means (calc class) the differentiation operator, or a tangent vector (diff geom) [which is again an operator on functions]. That is, "d/dx" is regarded as a single symbol, not a quotient of exterior derivative by a differential. > d^2 f/dx^2 is actually wrong as the way to write the second derivative ... I think this is true given your interpretation, based on your earlier computation. But again, in ordinary calc classes (which after all don't talk about differential forms) the "d^2 y/dx^2" notation comes from the standard calc class meaning for d/dx and it goes back ages. I just looked in G. H. Hardy's classic "Pure Mathematics" from over a hundred years ago and yep! - d^2 y /dx^2 is one of the notations for the second derivative. (I'm not sure if it goes back to Leibniz.) So if you're lobbying for a change you have your work cut out for you. :-) Regardless, I hadn't thought of interpreting "d/dx" as "exterior derivative divided by differential" - interesting idea. [edited: added note about d^2, minor edit in last sentence]
    - impossiblefork 3y ago
      It's clearly not exactly an exterior derivative. d(df) can't be seen as zero, since we will be dividing it by dx, and we will therefore need to retain things that would otherwise become infinitesimal.
      - bikenaga 3y ago
        If "d" isn't exterior derivative, I don't see how it's defined. I can't think of any other meaning "d" could have in the context of this discussion. Sorry, I'm just not understanding you - but thank you anyway for trying to explain your thinking.
        impossiblefork 3y ago
        It's some kind of formal analogue where you can't drop all terms that become infinitesimal, sort of like a formal power series compared to an actual power series. So something like an exterior derivative where d(df) is not defined to be zero and such terms must be kept.
  - bikenaga 3y ago
    One other point about the earlier computation. Besides d^2 being 0 if d is the exterior derivative, there's another difficulty. Differential forms are elements of an exterior algebra Λ(W) where W is (say) a real vector space. Addition and multiplication of forms are defined; division is not. So before you can write "1/dx" and have it mean "division by dx", you have to first define what you mean by division by a differential form. The definition of the operation precedes the use of the symbol. I'm not sure how you'd define division of differential forms - you're going to take two differential forms, divide, and get a differential form? -but by all means explain if you have something in mind. I still think it was a good thought to ask whether "d/dx" could be interpreted as "exterior derivative divided by dx".

The problem is that d/dx(dy/dx) is not d^2 y/(dx)^2. d/dx (dy/dx) is really d(dy/dx)/dx, which is ((d^2 y)/dx + dy d(1/dx))/dx = d^2 y/dx - dy/dx (d^2 x)/(dx)^2, where df is the actual differential.

bikenaga 3y ago

We're probably talking about different things. I'm not using "d/dx" as the differential operator "d" divided by the differential "dx", which is what I think you mean. I mean the compound symbol "d/dx" from calculus, which means differentiation with respect to x - or alternatively (from a differential geometric point of view) the tangent vector "d/dx" operating on the function f. I meant to explain why the notation "d^2 f/dx^2" is the way it is. It looks funny, and students would often ask why the 2's were "in different places". But it makes sense when you remember that "d/dx" is the derivative operator.

impossiblefork 3y ago
I am talking about d/dx as the differential operator d divided by dx. That's the way to make it actually work, algebraically. d^2 f/dx^2 is actually wrong as the way to write the second derivative unless df/dx = 0 or d^2 x/(dx)^2 = 0; the first case is trivial, and the second is almost never the case. Suppose for example that in actually x = t^2. Then dx = 2t dt, d(dx) = 2(dt)^2 + 2 t (d^2 t) and we get (2(dt)^2 + 2t(d^2 t))/(2t dt)^2 = 0.5 + (...) d^2 t/(dt)^2, so it certainly isn't zero. But if you have actual differentials you can do algebra.
- bikenaga 3y ago
  > I am talking about d/dx as the differential operator d divided by dx. Okay, that's what I figured from your earlier computation. (One note: If "d" means exterior differentiation, then d^2 = 0.) But if you mean d/dx in this way you should make sure you tell people, because that's not the interpretation it has in calculus (say a typical calc class in a typical college) or in differential geometry. In those cases, d/dx means (calc class) the differentiation operator, or a tangent vector (diff geom) [which is again an operator on functions]. That is, "d/dx" is regarded as a single symbol, not a quotient of exterior derivative by a differential. > d^2 f/dx^2 is actually wrong as the way to write the second derivative ... I think this is true given your interpretation, based on your earlier computation. But again, in ordinary calc classes (which after all don't talk about differential forms) the "d^2 y/dx^2" notation comes from the standard calc class meaning for d/dx and it goes back ages. I just looked in G. H. Hardy's classic "Pure Mathematics" from over a hundred years ago and yep! - d^2 y /dx^2 is one of the notations for the second derivative. (I'm not sure if it goes back to Leibniz.) So if you're lobbying for a change you have your work cut out for you. :-) Regardless, I hadn't thought of interpreting "d/dx" as "exterior derivative divided by differential" - interesting idea. [edited: added note about d^2, minor edit in last sentence]
  - impossiblefork 3y ago
    It's clearly not exactly an exterior derivative. d(df) can't be seen as zero, since we will be dividing it by dx, and we will therefore need to retain things that would otherwise become infinitesimal.
    - bikenaga 3y ago
      If "d" isn't exterior derivative, I don't see how it's defined. I can't think of any other meaning "d" could have in the context of this discussion. Sorry, I'm just not understanding you - but thank you anyway for trying to explain your thinking.
      - impossiblefork 3y ago
        It's some kind of formal analogue where you can't drop all terms that become infinitesimal, sort of like a formal power series compared to an actual power series. So something like an exterior derivative where d(df) is not defined to be zero and such terms must be kept.
- bikenaga 3y ago
  One other point about the earlier computation. Besides d^2 being 0 if d is the exterior derivative, there's another difficulty. Differential forms are elements of an exterior algebra Λ(W) where W is (say) a real vector space. Addition and multiplication of forms are defined; division is not. So before you can write "1/dx" and have it mean "division by dx", you have to first define what you mean by division by a differential form. The definition of the operation precedes the use of the symbol. I'm not sure how you'd define division of differential forms - you're going to take two differential forms, divide, and get a differential form? -but by all means explain if you have something in mind. I still think it was a good thought to ask whether "d/dx" could be interpreted as "exterior derivative divided by dx".