Date: Sun, 18 Jul 2000 02:57:22 -0400

From: Alex Bogomolny

- Never use the word gradient as a substitute for the derivative
- If you are really curious, get the book Calculus by Gilbert Strang.
- Here's not a proof but an explanation:

Imagine a piece-wise linear function that gets values f(1) at 1, f(2) at 2, and so on. Between 1 and 2 it's linear, just as it is between, 2 and 3, 3 and 4, and so on.

Calculate its derivative. The derivative does not exists at integer points 1, 2, 3, ... But, elsewhere it's a piece-wise constant function. It's f(2) - f(1) on the interval between 1 and 2, it's f(3) - f(2) on the interval between 2 and 3, and so on.

The integral of the derivative is the sum of the areas of rectangles with the unit base. Which makes it equal to

(f(2) - f(1)) + (f(3) - f(2)) + (f(4) - f(3)) + ... + (f(N) - f(N-1)) = f(N) - f(1)

This is actually is what you were looking for: integral of the derivative is f(N) minus a constant (f(1), here).

To understand what I said, it's best to draw the diagram with the graphs of the two functions. Do not make me feel that I wasted my time.

Best,

Alexander Bogomolny

64264550 |