4 ms·

Since this is mentioning d/dx stuff... In the past I've read that that stuff is more or less just a shorthand for the "formal" definition of a derivative. But

by rtpg 3y ago

Since this is mentioning d/dx stuff... In the past I've read that that stuff is more or less just a shorthand for the "formal" definition of a derivative. But I have also seen people solve problems through manipulation of the dx's and the dy's, and treading that as... well as a fraction. Is there some sort of guide to what you're "allowed" to do with dx and dy in general? Or is the thing basically that people are just "being careful"? This, like discussion of set theory, is always tough because I feel like I'm missing the axioms that could justify certain operations

bg46z 3y ago
d/dx is not a fraction but an operation, dy/dx is a fraction
- rtpg 3y ago
  What series of operations can lead you to end up with dy/dx though? Like Integral{x^2){dx} is how you're introducing dx into a formula, but it's not like integration just works over division, so I'm a bit stuck on the construction phase
  - lupire 3y ago
    y = x^2 dy = 2x dx dy/dx = 2x Chain rule makes this not completely trivial formulation.
- xvedejas 3y ago
  Is `dy/dx` ever distinct from `d/dx y`? Generally I've treated `df(x)/dx` as just another way to write the operator `d/dx` applied to `f(x)`, which I don't think of as a fraction.
  - lupire 3y ago
    dy/dt = dy/dx * dx/dt
    - xvedejas 3y ago
      That's just the chain rule, right? d/dt y(t) = ( d/dx y(x) )( d/dt x(t) )
baazaa 3y ago
I get the impression that this is a relic of how calculus developed. Basically the creators had a vague intuition of an infinitesimal, it was later replaced by supposedly much more rigorous limit definitions. Then it turns out just creating a non-standard system in which infinitesimals exist is perfectly fine, just like imaginary numbers are useful. The equivalent to i = sqrt(-1) might be something like the nilpotent infinitesimal where epsilon^2 = 0 but epsilon =/= 0.
duncan_britt 3y ago
I was told by my Calc 2 instructor that in the past, people treated dx like a variable, but in modern times, mathematicians have identified some edge cases which make that kind of thing problematic if mathematical rigor is your thing.
- msla 3y ago
  You can make it rigorous again by studying nonstandard analysis: https://en.wikipedia.org/wiki/Nonstandard_analysis https://en.wikipedia.org/wiki/Nonstandard_analysis
markisus 3y ago
Another commenter has mentioned nonstandard analysis but I find that most people (commonly physicists, engineers) who are manipulating dy and dx are not actually using the framework of nonstandard analysis. What usually is going on is you mentally replace dx and dy with Delta x and Delta y = y(x + Delta x) - y(x), where Delta x is some small nonzero quantity. After doing your manipulations, you take the limit as Delta x -> 0. Then you have rigorous statements like Delta y / Delta x -> f'(x) (this is the definition of derivative), and Delta x/Delta x = 1, and (Delta x)^2/Delta x -> 0 (ability to neglect second order terms), etc. You do this type of thing so many times and it becomes rote and annoying to explicitly mention the limits and the reader is trusted to formalize it themselves as an exercise. The stuff that you are "allowed" to do is taught in a first course in real analysis. After taking such a course, you will be able to justify for yourself which manipulations are valid.
mikebenfield 3y ago
Yes, it is possible to treat d as a differential operator, and then dy and dx are the differentials of functions, and it makes sense to take their ratio. You can learn about this by reading about differential forms: https://en.wikipedia.org/wiki/Differential_form https://en.wikipedia.org/wiki/Differential_form Unfortunately I don't know a treatment of this subject that explains how to apply this concept to basic calculus without introducing some other more difficult concepts.
- JBits 3y ago
  Differential forms looks really cool and I really want to learn more about them at some point! That said, isn't calculus, mainly the chain rule, enough to tell you what operations are allowed?
  - mikebenfield 3y ago
    I definitely didn't understand how to use differentials just by learning calculus, but maybe it is possible.
  - lupire 3y ago
    For single variable calculus (y=f(x)), yes.
mananaysiempre 3y ago
For scalar functions of a scalar variable, a less drastic option than nonstandard analysis (and perhaps one that’s more useful for bridging into more advanced stuff) can be as follows: For any expression y involving x [*], denote by dy the linear part of y|x+t - y|x (that is, y evaluated at x+t less y evaluated at x). For y=f(x), that’s just a fancy way of saying f'(x)t, of course, but the intent is that where y was an “x-dependent scalar”, dy is an “x-dependent linear function” (without a constant term, as it is the convention in most settings outside of high school). The baby version stops at that: by our rules, dx = t, df(x) = f'(x)t = f'(x)dx, f'(x) = df(x)/dx, the not-a-proof for the chain rule becomes an actual proof except everything is assumed differentiable (where a textbook one would only need differentiable inputs), etc. Of course, at this stage it seems slightly miraculous that nothing ends up t-dependent, but it is what it is. (If you want, you can imagine that the symbol t is “private” to the previous paragraph so it’s not allowed to escape to “the user”, but then you need to prove that it actually doesn’t.) The adolescent version unholsters linear algebra: While of course every one-dimensional (real) vector space (i.e. a line with a chosen zero point) is R in disguise, there are multiple choices for what the disguise is (differing by a multiplication by a constant), and you might not know which one to prefer. (This is what choosing a basis in a one-dimensional space amounts to.) Given such a space and two vectors u (whatever) and v (nonzero), denote by u/v the number such that u = (u/v)v (the coefficient of proportionality, aka the coordinate of v when e is the basis vector). Now you can prove, for example, that u/w = (u/v)(v/w), because it is so when you choose any (single-vector) basis and substitute for each vector its (only) coordinate. (Can you make sense of uv/vw? Yes, as u⊗v / v⊗w, but tensor products are their own can of worms and probably overkill at this point.) At each value of x, the space of linear functions (of the private variable t) is one-dimensional, so everything in the previous paragraph applies. When we write dy/dx and so on, we mean the things from there except we do them at each value of x (“pointwise”). One of the things that this more advanced thinking gets you is that you can imagine how all of it generalizes to multiple variables. (Writing v/e_i for coordinates of v in the basis (e_i) is not common, but it does not not make sense—as long as you remember you can only “divide” by a basis, not by a single vector. Write out the coordinate transformation rules in this notation. The differential version will have you end up with df(x)/dx_i instead of the more common ∂f(x)/∂x_i, but again, that makes sense in context—note that, once again, the partial derivative wrt one coordinate on a plane depends on what the other coordinates on that plane are!) The grown-up version just says I’ve been talking about the cotangent bundle in wishy-washy language. (An “x-dependent scalar”? What’s that? Does it taste good?) Hopefully it was still of some help. [*] People do write df/dx where I would require df(x)/dx, and it’s convenient to do that, but I’m trying to avoid additional abuses of notation where possible.

by rtpg 3y ago

bg46z 3y ago

d/dx is not a fraction but an operation, dy/dx is a fraction

rtpg 3y ago
What series of operations can lead you to end up with dy/dx though? Like Integral{x^2){dx} is how you're introducing dx into a formula, but it's not like integration just works over division, so I'm a bit stuck on the construction phase
- lupire 3y ago
  y = x^2 dy = 2x dx dy/dx = 2x Chain rule makes this not completely trivial formulation.
xvedejas 3y ago
Is `dy/dx` ever distinct from `d/dx y`? Generally I've treated `df(x)/dx` as just another way to write the operator `d/dx` applied to `f(x)`, which I don't think of as a fraction.
- lupire 3y ago
  dy/dt = dy/dx * dx/dt
  - xvedejas 3y ago
    That's just the chain rule, right? d/dt y(t) = ( d/dx y(x) )( d/dt x(t) )

baazaa 3y ago

I get the impression that this is a relic of how calculus developed. Basically the creators had a vague intuition of an infinitesimal, it was later replaced by supposedly much more rigorous limit definitions. Then it turns out just creating a non-standard system in which infinitesimals exist is perfectly fine, just like imaginary numbers are useful. The equivalent to i = sqrt(-1) might be something like the nilpotent infinitesimal where epsilon^2 = 0 but epsilon =/= 0.

duncan_britt 3y ago

I was told by my Calc 2 instructor that in the past, people treated dx like a variable, but in modern times, mathematicians have identified some edge cases which make that kind of thing problematic if mathematical rigor is your thing.

msla 3y ago
You can make it rigorous again by studying nonstandard analysis: https://en.wikipedia.org/wiki/Nonstandard_analysis https://en.wikipedia.org/wiki/Nonstandard_analysis

markisus 3y ago

Another commenter has mentioned nonstandard analysis but I find that most people (commonly physicists, engineers) who are manipulating dy and dx are not actually using the framework of nonstandard analysis. What usually is going on is you mentally replace dx and dy with Delta x and Delta y = y(x + Delta x) - y(x), where Delta x is some small nonzero quantity. After doing your manipulations, you take the limit as Delta x -> 0. Then you have rigorous statements like Delta y / Delta x -> f'(x) (this is the definition of derivative), and Delta x/Delta x = 1, and (Delta x)^2/Delta x -> 0 (ability to neglect second order terms), etc. You do this type of thing so many times and it becomes rote and annoying to explicitly mention the limits and the reader is trusted to formalize it themselves as an exercise. The stuff that you are "allowed" to do is taught in a first course in real analysis. After taking such a course, you will be able to justify for yourself which manipulations are valid.

mikebenfield 3y ago

Yes, it is possible to treat d as a differential operator, and then dy and dx are the differentials of functions, and it makes sense to take their ratio. You can learn about this by reading about differential forms: https://en.wikipedia.org/wiki/Differential_form https://en.wikipedia.org/wiki/Differential_form Unfortunately I don't know a treatment of this subject that explains how to apply this concept to basic calculus without introducing some other more difficult concepts.

JBits 3y ago
Differential forms looks really cool and I really want to learn more about them at some point! That said, isn't calculus, mainly the chain rule, enough to tell you what operations are allowed?
- mikebenfield 3y ago
  I definitely didn't understand how to use differentials just by learning calculus, but maybe it is possible.
- lupire 3y ago
  For single variable calculus (y=f(x)), yes.

mananaysiempre 3y ago

For scalar functions of a scalar variable, a less drastic option than nonstandard analysis (and perhaps one that’s more useful for bridging into more advanced stuff) can be as follows: For any expression y involving x [*], denote by dy the linear part of y|x+t - y|x (that is, y evaluated at x+t less y evaluated at x). For y=f(x), that’s just a fancy way of saying f'(x)t, of course, but the intent is that where y was an “x-dependent scalar”, dy is an “x-dependent linear function” (without a constant term, as it is the convention in most settings outside of high school). The baby version stops at that: by our rules, dx = t, df(x) = f'(x)t = f'(x)dx, f'(x) = df(x)/dx, the not-a-proof for the chain rule becomes an actual proof except everything is assumed differentiable (where a textbook one would only need differentiable inputs), etc. Of course, at this stage it seems slightly miraculous that nothing ends up t-dependent, but it is what it is. (If you want, you can imagine that the symbol t is “private” to the previous paragraph so it’s not allowed to escape to “the user”, but then you need to prove that it actually doesn’t.) The adolescent version unholsters linear algebra: While of course every one-dimensional (real) vector space (i.e. a line with a chosen zero point) is R in disguise, there are multiple choices for what the disguise is (differing by a multiplication by a constant), and you might not know which one to prefer. (This is what choosing a basis in a one-dimensional space amounts to.) Given such a space and two vectors u (whatever) and v (nonzero), denote by u/v the number such that u = (u/v)v (the coefficient of proportionality, aka the coordinate of v when e is the basis vector). Now you can prove, for example, that u/w = (u/v)(v/w), because it is so when you choose any (single-vector) basis and substitute for each vector its (only) coordinate. (Can you make sense of uv/vw? Yes, as u⊗v / v⊗w, but tensor products are their own can of worms and probably overkill at this point.) At each value of x, the space of linear functions (of the private variable t) is one-dimensional, so everything in the previous paragraph applies. When we write dy/dx and so on, we mean the things from there except we do them at each value of x (“pointwise”). One of the things that this more advanced thinking gets you is that you can imagine how all of it generalizes to multiple variables. (Writing v/e_i for coordinates of v in the basis (e_i) is not common, but it does not not make sense—as long as you remember you can only “divide” by a basis, not by a single vector. Write out the coordinate transformation rules in this notation. The differential version will have you end up with df(x)/dx_i instead of the more common ∂f(x)/∂x_i, but again, that makes sense in context—note that, once again, the partial derivative wrt one coordinate on a plane depends on what the other coordinates on that plane are!) The grown-up version just says I’ve been talking about the cotangent bundle in wishy-washy language. (An “x-dependent scalar”? What’s that? Does it taste good?) Hopefully it was still of some help. [*] People do write df/dx where I would require df(x)/dx, and it’s convenient to do that, but I’m trying to avoid additional abuses of notation where possible.