(Homework Help)
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Suppose that  f(x)  is continuously differentiable
 
Suppose that  f(x)  is continuously differentiable
 
on the interval  [a,b].  Let  N  be a positive integer
 
on the interval  [a,b].  Let  N  be a positive integer
and let M = Max { |f'(x)| : a <= x <= b }.  Let
+
and let <math> M = Max { |f'(x)| : a \leq x \leq b } </math>.  Let
h = (b-a)/N  and let  R_N  denote the "right endpoint"
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<math> h = \frac{(b-a)}{N} </math> and let  <math> R_N </math> denote the "right endpoint"
 
Riemann Sum for the integral
 
Riemann Sum for the integral
 
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<math> I = \int_a^b f(x) dx .</math>
I = int( f(x), x=a..b).
+
 
+
 
In other words,
 
In other words,
 +
<math> R_N = \sum_{n=1}^N f(a + n h) h .</math>
  
R_N = sum( f(a + n*h)*h , n=1..N ).
+
Explain why the error, <math> E = | R_N - I | </math>, satisfies
 
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<math> E < \frac{M(b-a)^2}{N}. </math>
Explain why the error, E = | R_N - I |, satisfies
+
 
+
E < M(b-a)^2/N.
+
 
</blockquote>
 
</blockquote>
  
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.
+
*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?
+
*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?
+
*Chumbert: Yeah, he said in class today (Wed.) to assume that, right?
  
 
==Interesting Articles about Calculus==
 
==Interesting Articles about Calculus==

Revision as of 08:13, 18 September 2008

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 \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 < \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?

Interesting Articles about Calculus

The minimum volume happens at the average_MA181Fall2008bell

Learn LaTeX_MA181Fall2008bell

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett