Please HELP Don't Understand Simple Concept on Riemann Sums

Click For Summary
SUMMARY

The discussion clarifies the concept of Riemann sums, specifically how the norm of a partition approaches zero as the number of subintervals increases to infinity. The norm, defined as the largest subinterval, decreases as the intervals become smaller, necessitating an increase in the number of intervals to maintain a constant area under the curve. An example using the function f(x) = x^2 on the interval [0, 1] illustrates this concept, demonstrating that as the number of rectangles increases, the Riemann sum converges to the definite integral of the function.

PREREQUISITES
  • Understanding of Riemann sums
  • Basic knowledge of calculus, specifically integration
  • Familiarity with limits and convergence
  • Ability to manipulate summations and functions
NEXT STEPS
  • Study the properties of Riemann integrals
  • Learn about the Fundamental Theorem of Calculus
  • Explore numerical integration techniques, such as the Trapezoidal Rule
  • Investigate the concept of convergence in sequences and series
USEFUL FOR

Students of calculus, educators teaching integration techniques, and anyone seeking to deepen their understanding of Riemann sums and their applications in mathematical analysis.

BuBbLeS01
Messages
602
Reaction score
0
Please HELP...Don't Understand Simple Concept on Riemann Sums

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??
 
Physics news on Phys.org
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 into 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.

I hope that made things clearer, even without pictures. Maybe this will help you visualize it.
 
Last edited:

Similar threads

  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K