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*Chumbert - Yeah, he said in class today (Wed.) to assume that, right? | *Chumbert - Yeah, he said in class today (Wed.) to assume that, right? | ||
| − | *Bell - Oops! Sorry about that. You're right. It needs to be <math>\le</math>. (I can show that the only time it is actually equal is when the function <math>f(x)</math> is a constant function.) | + | *Bell - Oops! Sorry about that. You're right. It needs to be <math>\le</math>. (I can show that the ''only'' time it is actually equal is when the function <math>f(x)</math> is a constant function.) |
*Ctuchek - I do remember him saying that we will need to use the Mean Value Theorem. | *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: | *Chumbert - Logically, I think I got it, but I'm not entirely sure how to prove it mathematically: | ||
Revision as of 07:10, 19 September 2008
Contents
Math 181 Honors Calculus
Getting started editing
Lecture Notes
Homework Help
Hello, this is gary from ma181. let's solve the extra credit problem. Here is the problem in italics:
Extra Credit Problem
Suppose that f(x) is continuously differentiable on the interval [a,b]. Let N be a positive integer and let $ M = Max { |f'(x)| : a \leq x \leq b } $. Let $ h = \frac{(b-a)}{N} $ and 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?
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