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On Leibniz Notation

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Re: On Leibniz Notation

#91

Post by bikenaga »

impossiblefork wrote: post #89
bikenaga wrote: post #82

Earlier quoted context omitted.

The second derivative notation makes sense when you think of d/dx as an operator on functions. (As someone noted below, d/dx is essentially a tangent vector, which is something that operates on functions.) d/dx takes the derivative of what follows with respect to x, so (d/dx)(f) = df/dx is the derivative function of f with respect to x. Now you want to differentiate df/dx with respect to x, so you do "d/dx" to "df/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.

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.

Re: On Leibniz Notation

#92

Post by bvan »

Buttons840 wrote: post #33

I've learned calculus a few times in and out of school. I learned it best from a home-schooling focused text book. The author published a paper describing his notation [0]; see section 3 which is very readable. The paper explains: > Most calculus students glaze over the notation for higher derivatives, and few, if any, books bother to give any reasons behind what the notation means. It's important to go back and cons…

Thanks for that reference. Very useful.

Re: On Leibniz Notation

#93

Post by bikenaga »

tolmasky wrote: post #64

Is there a good explanation of u substitution (or any other integral stuff that makes use of the denominator) using non-Liebnitz notation? Is it just that ∂1 is actually ∂/∂1, and thus you can begin using ∂1 the way you would have used ∂x?

If you mean u-substitution in definite integrals, that is something called the change of variables theorem. I think you mean indefinite integrals or antiderivatives, and I think you're asking for an explanation with using the "du/dx" sort of notation. Here's a quick explanation which may be more than you wanted to know. :-)

You have ∫ f(g(x)) g'(x) dx. [So it might be something like ∫ [cos(x^2)] (2 x) dx, where f(x) = cos x and g(x) = x^2.] You do the u-substitution u = g(x), so u'(x) = g'(x):

  ∫ f(g(x)) g'(x) dx = ∫ f(u) u'(x) dx
You want to say that the second integral is ∫ f(u) du -- essentially (to go back to "fractional" derivative notation), you want to do (du/dx) dx = du to get rid of the x's. Why is this justified?

Let F(u) = ∫ f(u) du: F is the antiderivative of f. By definition of the antiderivative, F'(u) = f(u). So by the Chain Rule,

  d/dx F(u) = F'(u) u'(x) = f(u) u'(x).
But since F(u) = ∫ f(u) du differentiates to f(u) u'(x), by definition F(u) is the antiderivative of f(u) u'(x). But the antiderivative of f(u) u'(x) is ∫ f(u) u'(x) dx. So putting all this stuff together,

  ∫ f(g(x)) g'(x) dx = ∫ f(u) u'(x) dx = F(u) = ∫ f(u) du.

Re: On Leibniz Notation

#94

Post by impossiblefork »

bikenaga wrote: post #91
impossiblefork wrote: post #89

Earlier quoted context omitted.

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.

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…

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.

Re: On Leibniz Notation

#95

Post by bikenaga »

impossiblefork wrote: post #94
bikenaga wrote: post #91

Earlier quoted context omitted.

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…

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…

> 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]

Re: On Leibniz Notation

#96

Post by impossiblefork »

bikenaga wrote: post #95
impossiblefork wrote: post #94

Earlier quoted context omitted.

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…

> 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/d…

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.

Re: On Leibniz Notation

#97

Post by bikenaga »

impossiblefork wrote: post #94
bikenaga wrote: post #91

Earlier quoted context omitted.

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…

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…

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".

Re: On Leibniz Notation

#98

Post by bikenaga »

impossiblefork wrote: post #96
bikenaga wrote: post #95

Earlier quoted context omitted.

> 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/d…

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.

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.

Re: On Leibniz Notation

#99

Post by impossiblefork »

bikenaga wrote: post #98
impossiblefork wrote: post #96

Earlier quoted context omitted.

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.

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.

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.

Re: On Leibniz Notation

#100

Post by ngcc_hk »

tobinfricke wrote: post #10

The discussion in Sussman and Wisdom's "Structure and Interpretation of Classical Mechanics", about how the Euler-Lagrange equations don't literally make sense as traditionally written, has long resonated with me: https://mitp-content-server.mit.edu/books/content/sectbyfn/b... They also adopt a notation where partial derivatives are taken with respect to "argument slots".

The page is quite dated now: https://mitp-content-server.mit.edu/books/content/sectbyfn/b....

Is the a recommended vim environment for running the scheme in that book? I am usual to vim and slimv via https://susam.net/blog/lisp-in-vim.html#get-started-with-sli....

Really not emacs. Racket environment is a bit confusing.

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