Suppose that $ f(x) $ is continuously differentiable on the interval [a,b].
- Let N be a positive integer
- Let $ M = Max \{ |f'(x)| : a \leq x \leq b \} $
- Let $ h = \frac{(b-a)}{N} $
- Let $ R_N $ denote the "right endpoint"
Riemann Sum for the integral
$ I = \int_a^b f(x) dx . $
In other words,
$ R_N = \sum_{n=1}^N f(a + n h) h . $
Explain why the error, $ E = | R_N - I | $, satisfies
$ E \le \frac{M(b-a)^2}{N}. $
- So what does this equation "E < M(b-a)^2/N" mean. This reads that the error is less than the Maximum value of the derivative of the function of x multiplied by the interval squared from x=a to x=b all divided by the total number of subintervals N.
- I don't understand why this must be true. Maybe I'm wrong, but if f(x) were a horizontal line, wouldn't E=0 and M(b-a)^2/N also be =0. That would mean it is a false statement that E < M(b-a)^2/N. Are we to assume that E <= M(b-a)^2/N?
- Chumbert - Yeah, he said in class today (Wed.) to assume that, right?
- Bell - Oops! Sorry about that. You're right. It needs to be $ \le $. (I can show that the only time it is actually equal is when the function $ f(x) $ is a constant function.)
- Ctuchek - I do remember him saying that we will need to use the Mean Value Theorem.
- Chumbert - Logically, I think I got it, but I'm not entirely sure how to prove it mathematically:
The $ M(b-a) $ gives the height of one section(slope=(y/x), so slope*x=y), where $ \frac{(b-a)}{N} $ gives the width, and when multiplied together, they give you a rectangle which, if you remember from class, is the error--take the R-sum, then stack the extra blocks on to of each other. Does anyone else remember that? Or should I explain it better?
- It's a bit easier to follow the discussion when one puts a signature after a comment. Just push the signature button in the edit page, or type two dashes followed by four ~, i.e. --~~~~, and your signature with the date will appear. --Mboutin 17:42, 19 September 2008 (UTC)
- Dryg - I get what you're saying Chumbert.. Yeah I remember the explanation of the stacking of blocks of the error from each sum. Unfortunately, I don't know how to prove it mathematically either --Idryg 14:30, 22 September 2008 (UTC)
- Somebody ought to be able to find the argument in their class notes. I sketched the argument in class one day. --Bell 12:07, 23 September 2008 (UTC)
- Here are the Error Notes.--Gbrizend
