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==Homework Help== | ==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 <= x <= b }. Let | ||
| + | h = (b-a)/N and let R_N denote the "right endpoint" | ||
| + | Riemann Sum for the integral | ||
| + | |||
| + | I = int( f(x), x=a..b). | ||
| + | |||
| + | In other words, | ||
| + | |||
| + | R_N = sum( f(a + n*h)*h , n=1..N ). | ||
| + | |||
| + | Explain why the error, E = | R_N - I |, satisfies | ||
| + | |||
| + | E < 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? | ||
==Interesting Articles about Calculus== | ==Interesting Articles about Calculus== | ||
Revision as of 12:05, 16 September 2008
Contents
Math 181 Honors Calculus
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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 <= x <= b }. Let h = (b-a)/N and let R_N denote the "right endpoint" Riemann Sum for the integral
I = int( f(x), x=a..b).
In other words,
R_N = sum( f(a + n*h)*h , n=1..N ).
Explain why the error, E = | R_N - I |, satisfies
E < 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?
Interesting Articles about Calculus
The minimum volume happens at the average_MA181Fall2008bell
