> An integral is the inner product of a function and a suitably-dimensioned unit.
Does that mean that ∫f(x)dx is f.(dx, dx, …) = f(x_0)·dx + f(x_1)·dx + f(x_2)·dx + … for all x in the domain?
Does that mean that ∫f(x)dx is f.(dx, dx, …) = f(x_0)·dx + f(x_1)·dx + f(x_2)·dx + … for all x in the domain?
'Projection' also has no intuitive (EDIT: I meant 'intrinsic') meaning; "inner product" is the same structure as "projection + norm" (subject to appropriate axioms). Anyway, I didn't mean to claim that the definition was arbitrary, but rather that there was no way to argue against it: definitions can't be wrong (at worst, they can be infelicitous, uninteresting, or uninhabited).
> Intuitively, sum(f_i * g_i).
I think rndn (https://news.ycombinator.com/item?id=9621422 )'s objection applies to this intuition: to get a reasonable approximation of the integral, you need a lot of sample points, and any sum that doesn't take into account the spacing of those sample points has a good chance of diverging. (Consider f = g = 1, so that the sum is just a count of the number of sample points!)
Once you write sum(f(x_i) * g(x_i) * (dx)_i), of course, this becomes just notation for (a sequence of) Riemann sums, whose limit is by definition the integral (for continuous functions).
Indeed, I don't understand how it could be otherwise. To multiply functions pointwise, you need to know their values at points. It seems to me that 'sampling' is a very good word to describe the process of evaluating a function at a lot of points.
> as could "with respect to x", but the Calculus Gods will smite any who think of dx as a sample of x
Indeed not! It is the spacing between sample points. That is, the `dx` in an integral literally stands for the "ghost of [the] departed quantity" `x_{i + 1} - x_i` (and, in an infinitesimal approach to calculus, it doesn't just stand for but literally is such a difference).
I think that that is what is meant, except that it's not clear what you mean by "for all `x` in the domain"—`x` occurs bound on both sides. Of course this interpretation requires that one understand it as a philosophy rather than a calculation; for example, as your explicit version points out, one really needs tag points spaced `dx` apart to define the inner product, and (absent infinitesimals) the result will be only an approximation to the true integral.
stephencanon (https://news.ycombinator.com/item?id=9620263) gives another interpretation that is unimpeachably mathematically correct, but (a) it is so nearly circular that I think it must be not what bandrami (https://news.ycombinator.com/item?id=9616961) meant, and (b) (perhaps more importantly) the unit there is built into the definition of the inner product itself, rather than being part of the second "inner multiplicand".
By that point (this course was "differential operators", 700-level stuff) we all had a decent intuition of what inner and outer products are. The prof's comment was that looking at f(x) as an infinite-dimensioned vector, there is a unit *-cube g(x)=1 of compatible dimensions that can produce an inner product f|g (I'm not going to hunt through my character map for the dot or the integral sign). That inner product is the same as Integral(f(x), dx). This was in analogy to the differential operator being the exterior ("wedge") product of a function and its field.
The real point of the definition was relating y' and Integral(y) to div y and grad y: in y' you're going from vectors to tensors, and in Integral(y) you're going from vectors to scalars. Or, Integral(y) is a projection of y on some unit cube, and y' is finding the function of which y is the projection on an appropriate unit cube.
No, though it is true that the integral with respect to x, as an operation, is the limit of that summation as dx approaches 0. But the point wasn't about any particular numerical or symbolic manipulation we could do (this was a graduate-level calculus class, after all; we all knew how to actually integrate things).
Generally, when you learn inner products of functions, you learn the definition
f·g = ∫(f(x)g(x))dx where the concatenation there of f(x) and g(x) represents scalar multiplication. We all know how integration works, so this becomes how we define inner products.
My professor's point was to reverse the primacy there. We have a sense from vector operations of what inner products are; that can inform our intuition of what an integral is. That is, rather than saying "I know how to do an integral so I can now take an inner product of two functions", say "I have an intuition of what an inner product is, that is, the projection of one vector onto another to form a scalar, and that should inform my intuition of what an integral is".
The larger motivation for the whole talk was introducing Clifford algebras and the symmetry between dot product generalized to inner product opposed by outer product on the one hand, and cross product generalized to wedge product opposed by interior product on the other hand.
And, just to finish the mathjerking, the whole point of the course was to get to:
(δΩ)∫ω = (Ω)∫dω
ie, the most general case of Stokes' theorem. But that takes a lot of sussing out of what the differential operator d actually is.
Well, a proper math education is something that many people can only dream of.