3 ms·

So ∫f(x)g(x)dx = (f.g)·dx?

by rndn 11y ago

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).

So ∫f(x)g(x)dx = (f.g)·dx?

by 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).