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?
- Yeah, he said in class today (Wed.) to assume that, right? --Chumbert
- 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.
- 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? --Chumbert
- 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)
- So is this problem officially solved now that we have the notes of both Professor Bell and Gbrizend? I wish I had had more time to actually contribute to this. Or have we not actually proven anything yet? It sure looks like the solution is in the notes. --Jhunsber
- I think so... With said notes and the connections made, I think we're golden =]. Of course, my opinion doesn't really count for anything.. so we'll wait for Professor Bell's approval. We could write out in logical steps how it connects, with fancy LaTeX and everything. Anyone want to volunteer? =P --Chumbert
- Well, since I haven't actually contributed anything yet, and therefore feel that I am not eligible for the extra credit, I definitely volunteer. I'll have time this week, I'm sure. It should be up by Saturday.Jhunsber
