Calculus for mathematicians (1997) [pdf]

'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).

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.

I would say a course on trigonometry usually covers (my experience): trigonometric functions exact value of them for the angles 30, 45, 60, 90 ... degrees Trigonometric formulas for the sum and difference of angles. A formuka for the double and the half angle. Law of sine and law of cosine Lots of relations derived from the Pythagoras theorem (sin^2+cos^=1) how to solve trigonometric equations

With all this, you are equipped to completely determine a triangle, knowing some of its and the length of some its sides. As as application, I was taught, how to measure heights and distances provided you can measure angles.

Thus, without trigonometry, it would be fairly hard to take a course on analytic geometry.

Now, how would the course be enhanced by introducing sine as the solution of an ODE?

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).

If you don't have calculus, you don't have anything like a delta-epsilon proof of continuity. But without calculus, you also don't know that you need it. So you just assume (correctly) that you can interpolate, and it works just like you expect, and life goes on.