1. Nov 11, 2007

### BuBbLeS01

Can someone please explain this to me...

The number of subintervals in a partition approaches infinity as the norm of the partition approaches 0. That is, ||Triangle|| approaches 0 implies that n approaches infinity.

I thought the the norm (||Triangle||) was just the largest subinterval of a partition, so how does it approach 0??

2. Nov 11, 2007

### CompuChip

That is because we make the intervals smaller and smaller (going to zero), which means we have to make the number of intervals go to infinity to cover the entire domain (that is, ||Triangle|| * #{Triangles} = constant). Let me give an example.

Suppose we have a function, f(x) = x^2 on the interval [0, 1] and we want to integrate it. At first, we approximate it by a square of width 1 and height f(1) = 1, which gives 1. This is of course very rough, so we divide the square into two. One rectangle goes from 0 to 1/2 with height f(1/2) = 1/4, and the other one from 1/2 to 1 with height 1. This gives for the area (1/2) * (1/4) + (1/2) * 1 = 1/8 + 1/2 = 5/8.
Now divide it into four parts, all rectangles with width 1/4 and as height the value of the function in the rightmost point of each interval, as in this image. The are will become something like
$$\sum_{n = 1}^N \frac{1}{N} \times f(n/N),$$
where the first term is the width of the rectangle and the second term is the height (draw a picture for yourself, I can't do it here).
Now we make N larger and larger, dividing [0, 1] up in to more and more intervals. The rectangles get smaller and smaller. The idea is of course that the sum converges to $$\int_0^1 x^2 \, dx$$.
You can calculate for yourself, that you get
$$\frac{1}{N^3} \sum_{n = 1}^N n^2 = \frac{\frac16 N (N + 1) (2 N + 1)}{N^3}$$
which converges to 2/6 = 1/3 as N goes to infinity, which is indeed the surface under the curve.