6 ms·

> 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)

by rndn 11y ago

> 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?

stephencanon 11y ago
Integral[f(x)g(x)dx] defines an inner product on a space of functions (glossing over exactly what functions) on an interval. So Integral[f(x) dx] is the inner product (f, 1), where "1" is the constant function g(x) = 1 on the interval.
- rndn 11y ago
  So ∫f(x)g(x)dx = (f.g)·dx?
  - stephencanon 11y ago
    No, f·g = ∫f(x)g(x)dx.
    - rndn 11y ago
      But don't you need to account for the width dx of the slices you sum up? I.e. wouldn't f.g = f(x_0)g(x_0) + f(x_1)g(x_1) + … over a real interval blow up?
      - JadeNB 11y ago
        The point is that f.g (or, as a mathematician or physicist might write it, ⟨f, g⟩ or ⟨f|g⟩) has no independent meaning, but must be defined; and one way to define it (for `L^2` functions, the only one compatible with the `L^2` norm) is as stephencanon did at https://news.ycombinator.com/item?id=9620263 https://news.ycombinator.com/item?id=9620263 .
        bandrami 11y ago
        Well, it's not just any arbitrary definition, it's the projection of f onto g. Intuitively, sum(f_i * g_i).
        JadeNB 11y ago
        > Well, it's not just any arbitrary definition, it's the projection of f onto g. '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 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).
        bandrami 11y ago
        Sampling has nothing to do with this. The inner product of f (x) and g (x) simply is the integral with respect to x of their pointwise scalar products (I suppose in a very broad sense "pointwise" could be seen as analogous to sampling, as could "with respect to x", but the Calculus Gods will smite any who think of dx as a sample of x). The "ah ha" moment was in seeing inner product as the fundamental operation and integration as derived from it.
        JadeNB 11y ago
        > I suppose in a very broad sense "pointwise" could be seen as analogous to sampling 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).
JadeNB 11y ago
> 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? 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 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 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".
- bandrami 11y ago
  No, stephencanon's comment is what I mean, and this may be an example of what my prof. meant that my students would not then be able to teach calculus. 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.
bandrami 11y ago
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? 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.
- rndn 11y ago
  > mathjerking Well, a proper math education is something that many people can only dream of.

by rndn 11y ago

stephencanon 11y ago

Integral[f(x)g(x)dx] defines an inner product on a space of functions (glossing over exactly what functions) on an interval. So Integral[f(x) dx] is the inner product (f, 1), where "1" is the constant function g(x) = 1 on the interval.

rndn 11y ago
So ∫f(x)g(x)dx = (f.g)·dx?
- stephencanon 11y ago
  No, f·g = ∫f(x)g(x)dx.
  - rndn 11y ago
    But don't you need to account for the width dx of the slices you sum up? I.e. wouldn't f.g = f(x_0)g(x_0) + f(x_1)g(x_1) + … over a real interval blow up?
    - JadeNB 11y ago
      The point is that f.g (or, as a mathematician or physicist might write it, ⟨f, g⟩ or ⟨f|g⟩) has no independent meaning, but must be defined; and one way to define it (for `L^2` functions, the only one compatible with the `L^2` norm) is as stephencanon did at https://news.ycombinator.com/item?id=9620263 https://news.ycombinator.com/item?id=9620263 .
      - bandrami 11y ago
        Well, it's not just any arbitrary definition, it's the projection of f onto g. Intuitively, sum(f_i * g_i).
        JadeNB 11y ago
        > Well, it's not just any arbitrary definition, it's the projection of f onto g. '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 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).
        bandrami 11y ago
        Sampling has nothing to do with this. The inner product of f (x) and g (x) simply is the integral with respect to x of their pointwise scalar products (I suppose in a very broad sense "pointwise" could be seen as analogous to sampling, as could "with respect to x", but the Calculus Gods will smite any who think of dx as a sample of x). The "ah ha" moment was in seeing inner product as the fundamental operation and integration as derived from it.
        JadeNB 11y ago
        > I suppose in a very broad sense "pointwise" could be seen as analogous to sampling 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).

JadeNB 11y ago

> 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? 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 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 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".

bandrami 11y ago
No, stephencanon's comment is what I mean, and this may be an example of what my prof. meant that my students would not then be able to teach calculus. 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.

bandrami 11y ago

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? 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.

rndn 11y ago
> mathjerking Well, a proper math education is something that many people can only dream of.