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Just a thought on calculus notation..

It's hard to explain so I'll just write an example as to what I'm wondering. This one is in the context of Newtonian mechanics and displacement/velocity/acceleration as a function of time. Why isn't notation like this used when one integrates twice to the same variable?

x⃗=∫v⃗ dt=∬a⃗ dtdt=∬a⃗ (dt)2 \vec{x} = \int \vec{v} \ dt = \iint \vec{a} \ dt dt = \iint \vec{a} \ (dt)^2

I have never seen anything like this before and I'm just curious as to why. I am aware of when this is done with two seperate variables such as:

∫(x2+y2) dxdy \int (x^2 + y^2) \ dx dy

and not to mention that this kind of notation is used:

ddx(dydx)=d2y(dx)2=d2ydx2 \frac{d}{dx} ( \frac{dy}{dx}) = \frac{d^2 y}{(dx)^2} = \frac{d^2 y}{dx^2}

So is there any reason why integrating twice with respect to the same variable isn't shown as in the example? Or maybe it is but I have just never seen it?

Might seem like a stupid question :tongue:
Original post by TheBBQ

So is there any reason why integrating twice with respect to the same variable isn't shown as in the example?


Maybe because you wouldn't use it very often? Usually you stop after one integration to evaluate the arbitrary constant, then integrate again.
Here's one possible reason.

When you see dxdydxdy, you actually see dx∧dydx\wedge dy. That's the wedge product, and it's.. complicated, but kind of intuitive. The intuition is exactly what you'd expect; if dx, dy are lengths then dx∧dydx\wedge dy is the "area-element" given by dx and dy.

Now dxdx is not "an infinitesimal change in x", in the machinery that all this happens in. It is a differential form living in the tangent space at some point of some manifold for which x is a coordinate. Why does any of what I've said matter? Well, differential forms are the elements of an exterior algebra (wiki that - you'll see why the wedge is rather complicated but non-the-less intuitive in low dimensions) and here's the fact we care about: dx∧dx=0dx \wedge dx =0. So you can't integrate this! (well yeah you can integrate 0 just fine, but yknow.. it's not integrating f twice!)

NB: In this perspective, ∫f(x)dx\int f(x) dx doesn't mean "integrate f(x) with respect to x"; it means "integrate the 1-form f(x)dxf(x)dx. So the (dx)^2 isn't worth writing because you'd be integrating zero, no matter what f was.

NB 2: Now yeah.. you can integrate a function with respect to x, twice... I'm no expert in this area, but I would think it's because you integrate something different the second time. So first you integrate f(x)dxf(x)dx and get F(x)F(x). Now you make the new 1-form F(x)dxF(x)dx and you can then integrate that.

Added: Yeah, thinking more it's almost certainly that reason. Areas spanned by dx and dx are degenerate and integrating them won't give anything. You'd have to integrate twice separately along the length to not get just 0.



Added even more: Here's a fantastic blog post about this perspective. http://math.blogoverflow.com/2014/11/03/more-than-infinitesimal-what-is-dx/
(edited 8 years ago)

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